title videovideo introduction by prof. christopher lamoureux department of finance university of...

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Differentiation, P. & A.Differentiation. Properties & Applications: page 5(material continues)So far we have used differentiation for simple numerical problems and as new terminology for marginal analysis. Many other types of business problems can be solved with differentiation. As an illustration, we will apply differentiation to a situation in inventory control.Example 5. A large computer manufacturer uses 50,000 CD drives at a uniform rate throughout the year. These are outsourced, and must be ordered in batches from the supplier. Drives cost $20 each, and there is an additional $1,200 expense per order for processing, shipping, receiving and handling the drives. Thus, it is cheaper to order in large batches.The drives must be kept in inventory from the time that they arrive until they are used in computers. The estimated cost of keeping each drive stored is one cent per day.The cost of maintaining the inventory suggests placing smaller orders, that will be used up more quickly. Assuming that orders can be placed so that the drives arrive just in time to be used, how often should the company order a batch of drives, how many should be ordered, and how much will it cost them, per year, to order and store the CD drives?ITC93This is read as the derivative of f, at x, is f prime of x. The process of computing f (x) is called differentiation and f (x) is interpreted as the instantaneous rate of change in f(x) with respect to x.

Differentiation, DerivativesIn the new terminology, marginal demand, MD(q), is the derivative of the demand function, D(q). Hence, MD(q) = D(q). Likewise, MC(q) = C(q), MR(q) = R(q), and MP(q) = P(q). In this notation, the units of the marginal functions will reflect the units of the original functions.We can compute the approximate value of a derivative, f (x), byhaving Excel compute for a very small value of h.

Example 2. Evaluate f (2), if f(x) = x3 + 6. We let h = 0.00001and approximate f (2) with the difference quotient

Differentiation.Derivatives: page 2(material continues)Click here for technical information about our definition.ITC80

The And function is used as the statement in the IF function, completing the definition of f in Cell C9.Cell C9 in now dragged down to Cell C509.Graphing, Points(material continues)Graphing Functions. Plotting Points: page 4ITGraph Examples.xlsxC5Graphing, Points

The best graph of our 501 points is made using an XY (Scatter) plot, with smoothed lines and no markers. To find this, click Insert, then on the Create Chart boxWe select this type of plot and enter our data range.

Create Chart(material continues)Graphing Functions.Plotting Points: page 5ITGraph Examples.xlsxC6Graphing, Points

Right click in the plot region, then left click on Select Data. Once the data is entered, we can format the plot in any way that we like.(material continues)Graphing Functions. Plotting Points: page 6ITGraph Examples.xlsxC7Graphing, Points

The best effect is usually obtained by clicking on the plotted points, clicking on Format/ Selected Data Series, and then selecting a medium line width with no markers.Graphing Functions.Plotting Points: page 7Graph Examples.xlsxIT(material continues)C8Using Solver, Excels Solver1. EXCELS SOLVERThe utility Solver is one of Excels most useful tools for business analysis. This allows us to maximize, minimize, or find a predetermined value for the contents of a given cell by changing the values in other cells. Moreover, this can be done in such a way that it satisfies extra constraints that we might wish to impose. Example 1. The size limitations on boxes shipped by your plant are as follows. (i) Their circumference is at most 100 inches. (ii) The sum of their dimensions is at most 120 inches. You would like to know the dimensions of such a box that has the largest possible volume. Let H, W, and L be the height, width, and length of a box; respectively; measured in inches. We wish to maximize the volume of the box, V = HWL, subject to the limitations that the circumference C = 2H + 2W 100 and the sum S = H + W + L 120.This problem is set up in the Excel file Shipping.xlsx. We will outline its solution with screen captures and directions. First, enter any reasonable values for the dimensions of the box in Cells B7:D7.Shipping.xlsxUsing Solver. Excels Solver(material continues)ITC114Integration, Area Under A Curve1. AREA UNDER A CURVEThere is a very convenient way to picture the revenue that results from selling q items with a demand function D for a good. Since R(q) is the product of q and D(q), it is the area of a rectangle that is q units wide and D(q) dollars high. It is easy to picture such a rectangle under the graph of the demand function.(material continues)Integration. Area Under A CurveI

qD(q)Demand FunctionRevenueqD(q)TC136

Differentiation, P. & A.(material continues)Differentiation. Properties & Applications: page 9This expression for C(t) is entered into Differentiating.xlsm as f(x). The resulting plots are shown below. Large scale plots show that f(x) has only one possible minimum value, and that the minimum seems to occur when x is close to 40.ITDifferentiating.xlsmC97Marketing Example, Double Pricing1. DOUBLE PRICING In the buffalo steak dinner example that we have been following, Solver can be used in the computation cells of the sheet Solver in Dinners.xlsm. We find that D(4,014) = 0. Hence, the total possible revenue is

We have seen that the restaurant chain would obtain a maximum weekly profit of $14,052 by selling 2,025 dinners, priced at $22.21 each. Under this plan, the mangers note that 2,025 people are buying the dinners for less than they would have been willing to pay and that 4,014 2,025 = 1,989 people who would have been willing to buy dinners at a lower price are not paying money to the restaurants. Looking at the situation from another point of view, $79,999 2,025$22.21 $35,024 of potential revenue is being lost. How might the restaurants collect some of this money and increase their profit?

Marketing Example. Double Pricing(material continues)ITC203Graphing, Several

The formulas for f and g are entered in Cells C9 and D9, using the IF function for g. We select an XY (Scatter) plot, with smoothed lines and no markers. Right click in the plot region, then left click on Select Data. Enter the indicated data range.(material continues)Graphing Functions. Several Functions: page 2Graph Examples.xlsxITC13Graphing, Several

Click on Add under Legend Entries (Series). We name Series 1 as f and enter its data ranges. Series 2 is named g, and its data is entered. Finally, the plot is formatted as desired.(material continues)Graphing Functions. Several Functions: page 3ITGraph Examples.xlsxC14

Use the plot of f and g from Example 2 to estimate the two values of x at which f(x) = g(x). Exercise 7Graphing, SeveralLook at the graph in the sheet Graph 2 of Graph Examples.xlsx to see how the final formatting was done. The graph is copied on this page.(material continues)Graphing Functions. Several Functions: page 4ITGraph Examples.xlsxC15Title

Video introduction byProf. Christopher Lamoureux Department of FinanceUniversity of Arizona.Graphing FunctionsTrend LinesDemand, Revenue, Cost, and ProfitDifferentiationUsing SolverIntegrationMarketing Example 2009 by The Arizona Board of Regents for The University of Arizona. All rights reserved.CIMathematics forBusiness DecisionsPart 2Release 2, 2009MarketingComputer DrivesProject 1The Mathematical Association of America

Published and Distributed byPart 2: Calculus & Optimization

1Graphing, A Graphing UtilityGraphing.xlsm(material continues)Graphing Functions.A Graphing UtilityClick here for information on running the macro in Graphing.xlsm.3. A GRAPHING UTILITYIt is important that you know how to create your own graphs. This will be necessary for special plotting needs. Understanding the process of computer graphing also helps you know what the computer is doing and assists you in your interpretation of the results. However, in routine situations the creation of plots can be time consuming and could distract us from our main goal of understanding the business applications of a function. For this reason, we are supplying an Excel file, Graphing.xlsm, that serves as a graphing utility.To use Graphing.xlsm, open the file and follow the directions in the blue box. Be sure to note that you are to enter a formula for f(x) in terms of the letter x, not a cell reference. x is the only letter that can be used as a variable in Graphing.xlsm.ITC17Before running Graphing.xlsm, be sure that Excel is set for Automatic Calculation.When entering a function in Graphing.xlsm, you can use the letters s, t, u, v, or w as constants. If this is done, you must enter values for any letters that you use.For example, to plot f(x) = x s, where s is a number that you enter in Cell L17, enter the following function in Cell C18.= x^sAccept the function and then click on the Graph button at the right of the plot. Entering a new number in Cell L17 will automatically change the function and redraw the plot.If values are entered in Cells L17:L21; then constants s, t, u, v, or w may also be used to describe the range, [a, b], of the plot. If these constant names are used, they must be entered in Cells H18:I18 preceded with "=" signs. For example, =s.Graphing, UtilityGraphing Functions. A Graphing Utility: page 2IT(material continues)Graphing.xlsmC18 Use Graphing.xlsm to plot the graph of fX over the

interval [-5, 15]. Note that this function

is the p.d.f. for an exponential random variable with parameter = 3.5.

Use Graphing.xlsm to plot the graph of f(x) = 2,000e0.06t over the interval [-12, 12].Exercise 11Graphing, UtilityGraphing Functions. A Graphing Utility: page 3 Use Graphing.xlsm to plot the graph of F(x) over the

interval [-5, 15]. Note that this function

is the c.d.f. for an exponential random variable with parameter = 3.5.

Exercise 12(material continues)Exercise 13ITGraphing.xlsmC19 Let = 1, and use Graphing.xlsm to plot thegraph of over the interval [4, 8]. To enter the

constant in Excel, type PI(). In Excel, the square root function is entered as SQRT.

Redo Exercise 13, but leave the parameter as a constant. Changing and recomputing the sheet should change the plot correspondingly. Hint: use the built-in constant, s, as a replacement for .Exercise 14Graphing, UtilityGraphing Functions. A Graphing Utility: page 4(material ends)Exercise 15 (i) Redo Exercise 15, but leave the parameter as a constant. Changing and recomputing the sheet should change the plot correspondingly. Hint: use the built-in constant, s, as a replacement for . (ii) Display graphs for = 0, 1, 2, and 4.Exercise 16

Graphing.xlsmITC20Trend Lines, ModelsTrend Lines. Fitting Models: page 2(material continues)Case Load.xlsxITWe would like to find a formula for the function that represents the underlying pattern of the case load growth. To do this, we must first use our judgment to select the type of functions that we will consider. For the CPS data we note the following. (i) The data points appear to increase and curve slightly upward. (ii) Many changing populations have patterns of exponential growth. (iii) Experience in public administration around the country suggests that social service case loads increase exponentially. For these reasons we will assume that the number of cases handled in a year that is t years after 1998, is given by an exponential function of the form f(t) = uevt, where u and v are constants. Recall that e is a constant that is the base for the natural exponential function. It can be shown that e 2.71828. The exponential function is defined for all values of x, and is entered in Excel with upper case letters as EXP().Once we have selected a form for f(t), Excel can find values for the constants u and v that produce the function of the given type, that best fits the data points. Such a function is called a trend line.C22Trend Lines, Models(material continues)ITWe will select an exponential trend line for the CPS data. Open Case Load.xlsx to see a plot of the nine original data points.To add a trend line:1. Left click on any data point.2. Right click to see the pull-down menu.3. Click on Add Trendline.Trend Lines. Fitting Models: page 3

Case Load.xlsxC23Trend Lines, Models

5. Click onTrendline Options.4. Click on Exponential.Trend Lines. Fitting Models: page 4(material continues)ITCase Load.xlsxC24Trend Lines, Models

6. To move to 2010, set Forward to 4.7. Click on Display Equation on chart.8. Click on Close.Trend Lines. Fitting Models: page 5(material continues)ITCase Load.xlsxC25Trend Lines, Models(material continues)IT

The equation of the exponential trend line is displayed on the plot. We see that the coefficients are u = 2373.6435 and v = 0.1217. To use these numbers in the Excel file, you must either copy them manually, or with copy/paste.

Trend Lines. Fitting Models: page 6Case Load.xlsxC26Trend Lines, ModelsTrend Lines. Fitting Models: page 7(material continues)IT The choice of a basic form for a trend line is of major importance. Further training in business will help you recognize the appropriate types of functions to use in common situations.Important!Let f(t) = 2,373.6435e0.1217t. Since we are using t as the number of years since 1998, the expected case load in a year y is f(y 1998). Case Load.xlsx shows that the agency can expect 10,225 cases in the year 2010. Their load has been growing at a yearly rate of approximately 12%.The extension of a trend line beyond the range of the known data is called extrapolation. This must be done with great caution. It is not unreasonable to expect the current pattern of growth to continue for the next four years. However, it would be absurd to expect the same circumstances to continue for the next 100 years. As Case Load.xlsx shows, f(2106 1998) = 1,212,295,365. Obviously, we should not assume that CPS will handle over one billion cases in the year 2106.Case Load.xlsxC27Trend Lines, Models

Example 2. (A Linear Model) We will illustrate the effect of different types of trend lines by fitting a linear function to the CPS data points. This means looking for the best fitting curve of the formf(t) = at + b.This is done in exactly the same way that we used for the exponential model, except that we select the Linear box on Excels menu.Once the coefficients of the trend line are computed, we can click on the equation box and format the numbers. For the linear trend line, we have displayed a and b as integers.Trend Lines. Fitting Models: page 8(material continues)ITCase Load.xlsxC28Trend Lines, Models(material continues)IT

We see that the coefficients are a = 487 and b = 2,112. To use these numbers in the Excel file, you must either copy them manually, or with copy/paste.Trend Lines. Fitting Models: page 9Case Load.xlsxC29Let f(t) = 487t + 2,112. The expected case load in a year y is f(y 1998). Case Load.xlsx shows that the agency can expect 7,956 cases in the year 2010. This is approximately 22% fewer cases than were predicted with the exponential model. Will growth be exponential, linear, or neither? Only time can tell with any certainty.The difference in choice of models is particularly noticeable over long periods of extrapolation. As computed in Case Load.xlsx, the linear model predicts a load of 54,708 cases in the year 2106. This is approximately 0.005% of the 1,212,295,365 cases predicted by theexponential model!

Trend Lines, ModelsTrend Lines. Fitting Models: page 10(material continues)I

How do I know what type of trend line to use?Ask a tutorTCase Load.xlsxC30Trend Lines, ModelsTrend Lines. Fitting Models: page 11 (i) Use the exponential model in the CPS example to predict the case load in the year 2011. (ii) Repeat Part i, using the linear model.Exercise 1 (i) Fit a 3rd degree polynomial trend line through the data points in the CPS example. (ii) What case load does this model predict for the year 2010?Exercise 3 (i) Use the exponential model in the CPS example to predict the case load in the year 2013. (ii) Repeat Part i, using the linear model.Exercise 2 (i) Fit a 6th degree polynomial trend line through the data points in the CPS example. (ii) What case load does this model predict for the year 2010? (iii) Does increasing the degree of a polynomial trend line seem to improve its predictive value?Exercise 4(material continues)ITCase Load.xlsxC31Trend Lines, Models During the first 6 days of a special promotion, a small business records the following sales information. The dollar amount listed for each day is the total cumulative sales from the start of the promotion.

(i) Fit a Power trend line to the data and use it to estimate the total sales during the first 9 days of the promotion. (ii) Repeat Part i, using a Linear trend line. (iii) Which model do you think is more realistic?Exercise 5

Trend Lines. Fitting Models: page 12 Fit an exponential trend line to the following data and use the lines formula to predict the highest closing price during July.

Exercise 6

(material continues)IExcelTC32Trend Lines, Models (i) Fit a linear trend line to the data in Exercise 6, and use this model to predict the highest closing price during July. (ii) Fit a 3rd degree polynomial trend line to the data in Exercise 6, and use this model to predict the highest closing price during July. (iii) Do you think that trend lines can provide a reliable way to predict stock prices?Exercise 7Trend Lines. Fitting Models: page 13 Your company, which manufactures and distributes beach wear, has kept sales records for the months of March, April, May, June, and July of the current year. These indicate that sales volume is increasing. Which, if any, of the types of trend lines that are available in Excel might be used to predict sales volume for the rest of the year? Explain your answer.Exercise 8(material continues)ITExcelC33Trend Lines, ModelsTrend Lines. Fitting Models: page 14(material ends)I The Flimsy Plastic company produces toys for children in the age group from 3 to 6 years. A regional sales office has kept records on the number of children in this range that live in its territory.

(i) Experiment with several different types of trend lines to determine which model best fits the data. (ii) Use the type of trend line that you select to predict the number of 3-6 year olds that will be in the sales territory in the year 2011.Exercise 9T3-6 Year Olds in Sales TerritoryYear19971998199920002001Population (Ks)390410440460470Year20022003200420052006Population (Ks)510540570580620ExcelC34S-T (continued)Standardized random variable 2-82Symbol font 77

UUniform random variable 2-27

V-W-X-Y-ZVariance, continuous random variable 2-75 finite random variable 2-69 sample 2-85 sample mean 2-92IndexIndex: page 5A-D E-L M-P Q-TTC226D, R, C, & P, IncomeDemand, Revenue, Cost, & Profit. Income: page 2(material continues)Define the demand function to be D(q) = aq2 + bq + c, where a = 0.0000018, b = 0.0002953, and c = 30.19.

ITDinners.xlsmC36D, R, C, & P, IncomeThe amount of money that a producer receives from the sale of q units of its product is called revenue, and is denoted by the revenue function R(q). If D(q) is the demand function, and revenue will come from selling q units at a price of D(q) dollars per unit, then R(q) = qD(q).In the case of the restaurants, the amount of money that diners pay for q of the buffalo steak dinners per week is weekly revenue. The following graph is copied from the sheet Income in Dinners.xlsm.Setting a high price for the dinners will result in very few orders, and a small revenue. Setting a low price for the dinners will also bring in low revenue.(material continues)Demand, Revenue, Cost, & Profit. Income: page 3Dinners.xlsm

ITC37D, R, C, & P, IncomeIt appears from the graph that the maximum revenue would result from selling approximately 2,300 buffalo steak dinners. Looking back at the graph of the demand function, we see that a price of around $20 can be expected to result in the sale of 2,300 dinners. Open the sheet Income in Dinners.xlsm and explore this with the Computation cells.Demand, Revenue, Cost, & Profit. Income: page 4 (i) Find the price, per dinner, that will result in the sale of 2,600 buffalo steak dinners. (ii) What revenue can be expected from 2,600 dinners?Exercise 1(material continues)

ITDinners.xlsmC38 Suppose that D(q) = 0.00006q2 + 250 is the demand function for a certain model of audio speaker. (i) Use Graphing.xlsm to plot D(q) and the revenue function, R(q). (ii) Estimate the price that will yield the greatest revenue.Exercise 4 (i) Experiment with different values in Cell B19 of the sheet Income in Dinners.xlsm to find the number of dinners that would be sold at a price of $19.95. (ii) What revenue can be expected if dinners are priced at $19.95?Exercise 2 Suppose that the demand function for a good is given by D(q) = 0.1q + 150. (i) Use Graphing.xlsm to plot D(q) and the revenue function, R(q). (ii) Estimate the price that will yield the greatest revenue.Exercise 3D, R, C, & P, IncomeGraphing.xlsm(material continues)Demand, Revenue, Cost, & Profit. Income: page 5ITC39 The demand function for a new type of car alarm system is shown in the adjacent plot. Note that the quantities sold are given in thousands. Estimate revenue, in dollars, from selling 300,000 alarms.Exercise 5 (i) Use your teams data and Trend Lines to find the formula for a quadratic demand function. (ii) Plot D(q) and R(q) for your teams product. Use the same units as in the work on the Class Project in the sheet Functions of Marketing Focus.xlsm.Exercise 6D, R, C, & P, Income(material continues)Demand, Revenue, Cost, & Profit. Income: page 6I

ExcelTC40D, R, C, & P, Expenses & Profit2. EXPENSES AND PROFITEvery producer incurs costs in the production of its product. These usually include fixed costs, which do not depend upon the amount of a good that is produced, and variable costs. Fixed costs might include such things as plant overhead, minimal labor costs, and debt service. Variable (production) costs cover such items as materials, labor, and distribution expenses. We will denote fixed cost by C0, and denote the variable cost for q units of a good by VC(q).A producers total cost function, C(q), for the production of q units is given by C(q) = C0 + VC(q).Example 2. We will continue Example 1, and develop a cost function for the restaurant chains new buffalo steak dinners. Managers estimate that the new menu item will have to support $9,000 out of the complete weekly fixed cost of operating the chain. Hence, C0 = $9,000. The head chef makes the following estimates for the costs of preparing various numbers of dinners.Demand, Revenue, Cost, & Profit. Expenses & Profit(material continues)ITC41D, R, C, & P, Expenses & ProfitNotice that there is some economy of scale. The cost of preparing the first thousand dinners is $14,000, of preparing the second thousand dinners is an additional $8,000, and of preparing the third thousand dinners is an additional $6,000. In order to develop a formula for the variable cost function, we will have Excel fit a trend line through the three estimated data points. The managers know that, in former menu offerings, food preparation costs have usually turned out to follow power function models. That is, VC(q) = uqv, for some constants u and v. In the sheet Profit of the file Dinners.xlsm, we have plotted the data points and added a power trend line. Excel finds that u = 177 and v = 0.633. Thus, VC(q) = 177q0.663, and the total weekly cost function for the buffalo steak dinners is C(q) = C0 + VC(q) = 9,000 + 177q0.633.(material continues)Demand, Revenue, Cost, & Profit. Expenses & Profit: page 2Dinners.xlsm

ITC42D, R, C, & P, Expenses & Profit(material continues)

Demand, Revenue, Cost, & Profit.Expenses & Profit: page 3ITDinners.xlsmC43D, R, C, & P, Expenses & ProfitThe sheet Profit of Dinners.xlsm shows a plot of both the revenue and cost functions on a single set of axes. As the graph shows, revenue appears to exceed cost in a range of approximately 700 to 3,100 dinners per week. The sheet Profit also plots the profit function, P(q). A maximum profit of roughly $14,000 seems to result from the preparation and sale of approximately 2,000 dinners per week. Note that this is somewhat less than our estimate of 2,300 dinners per week that would produce the maximum revenue for the chain. Open the sheet Profit and explore this with the Computation cells. The producer of any good is interested in profit. We let P(q) be the profit obtained from producing and selling q units of a good at the price D(q).Profit = Revenue CostP(q) = R(q) C(q)(material continues)Demand, Revenue, Cost, & Profit.Expenses & Profit: page 4I

How can I recog-nize the graphs of demand, cost, and profit functions?Ask a tutorTDinners.xlsmC44

D, R, C, & P, Expenses & ProfitIDemand, Revenue, Cost, & Profit. Expenses & Profit: page 5(material continues)TDinners.xlsmC45D, R, C, & P, Expenses & ProfitDemand, Revenue, Cost, & Profit. Expenses & Profit: page 6 (i) Find the cost of preparing 2,300 buffalo steak dinners. (ii) What revenue can be expected from preparing and selling 2,300 dinners? (iii) What weekly profit can be expected from preparing and selling 2,300 dinners?Exercise 7 (i) Find the cost of preparing 1,500 buffalo steak dinners. (ii) What revenue can be expected from preparing and selling 1,500 dinners? (iii) What weekly profit can be expected from preparing and selling 1,500 dinners?Exercise 8 (i) Experiment with computation in the sheet Profit of Dinners.xlsm to find the number of dinners that could be sold at a price of $18. (ii) Find the cost of preparing these dinners. (iii) What profit can be expected from preparing and selling buffalo steak dinners at $18?Exercise 9(material continues)ITDinners.xlsmC46D, R, C, & P, Expenses & ProfitReturning to our plot of the demand function, D(q), we can estimate that a price of approximately $22 would produce sales of 2,000 dinners. In summary, the restaurant chain should prepare around 2,000 buffalo steak dinners per week and price them at approximately $22. If this is done, they can expect a weekly profit of around $14,000 from the new menu item.The graph of P(q) shows us how sensitive the weekly profit is to deviation from the optimal price per dinner of around $22. For example, using rough estimates from the plot of the demand function, it appears that raising the price of a dinner to $25 would reduce the demand to around 1,600 dinners per week. Estimating from the plot of the profit function, we see that this would reduce the chains weekly profit to around $13,000. That is a 7% drop from the maximum profit of approximately $14,000.

(material continues)Demand, Revenue, Cost, & Profit. Expenses & Profit: page 7ITDinners.xlsmC47 Estimate the percentage drop in weekly profit that would result from dropping the price of buffalo steak dinners from $22 to $20.Exercise 10D, R, C, & P, Expenses & ProfitDemand, Revenue, Cost, & Profit. Expenses & Profit: page 8 Use the graphs in this section to estimate the range of dinner prices that would yield weekly profits of at least $10,000.Exercise 11 Use the graphs in this section to estimate the range of dinner prices that would yield weekly revenues of at least $35,000.Exercise 12 Use the graphs in this section to estimate the weekly profit that would result from spending $30,000 to prepare buffalo steak dinners.Exercise 13(material continues)ITDinners.xlsmC48D, R, C, & P, Expenses & ProfitDemand, Revenue, Cost, & Profit. Expenses & Profit: page 9Excel(material continues)Graphing.xlsm Refer to the demand function given in Exercise 4. Suppose that the fixed cost for producing the speakers is C0 = $60,000 and that it costs $110 to produce each speaker. (i) Use Graphing.xlsm to plot D(q) and P(q). (ii) Use your graphs to estimate the number of speakers that should be produced, and how they should be priced in order to attain the maximum profit. (iii) Approximately what maximum profit might be expected?Exercise 15 Refer to the demand function given in Exercise 3. Suppose that the fixed cost for producing this good is C0 = $12,000 and that the variable costs are given by (i) Use Graphing.xlsm to plot D(q) and P(q). (ii) Use your graphs to estimate the number of units that should be produced, and how they should be priced in order to attain the maximum profit. (iii) Approximately what maximum profit might be expected?Exercise 14

ITC49 Refer to the demand function whose graph is shown in Exercise 5. The cost function for the alarms is plotted in the adjacent graph. Note that the quantities of alarms are given in thousands and the total costs are given in millions of dollars. (i) Estimate the number of alarms that would be sold at a price of $250 each. (ii) Estimate the revenue, in dollars, that would result from the sale of alarms priced at $250. (iii) Estimate the cost, in dollars, of producing the alarms that would be sold at $250. (iv) Estimate the profit, in dollars, that would result from the sale of alarms at $250.Exercise 16D, R, C, & P, Expenses & ProfitDemand, Revenue, Cost, & Profit.Expenses & Profit: page 10

(material continues)ITC50D, R, C, & P, Expenses & ProfitDemand, Revenue, Cost, & Profit. Expenses & Profit: page 11 (i) Use your teams data to find a formula for the cost function of your product. (ii) Plot C(q) and P(q) for your teams data. Use the same units as in the work on the Class Project in the sheet Functions of Marketing Focus.xlsm. Study the Focus pages to see how to use the custom programmed function COST.Exercise 17Excel(material continues)SALEEverything Must Go!ITC51D, R, C, & P, FocusHow can demand, revenue,cost, and profit functions help us price12-GB drives?Obviously, Card Tech would like to price its new drives in such a way that profit is maximized. To compute values of their profit function, they must first develop demand, revenue, and cost functions.

on the projectCalculus, Mathematics, Tests, Homework, Computers

MarketingComputerDrivesI(material continues)Class ProjectT

CMarketing Focus.xlsm52D, R, C, & P, Focus


MarketingComputerDrivesOur information on test markets and variable costs is summarized in the sheet Data of the Excel file Marketing Focus.xlsm. Since we are dealing with quite large numbers, we will adopt some conventions for units. Prices for individual drives are given in dollars. Revenues from sales in the national market are given in millions of dollars. Quantities of drives in the test markets are actual numbers of drives. Quantities of drives in the national market are given in thousands of drives.At the top of the sheet Functions in Marketing Focus.xlsm, these units are applied to our data, and the projected yearly sales in the national market are computed for each test market price.IT(material continues)Class ProjectCMarketing Focus.xlsm53


MarketingComputerDrivesSince it was assumed that the demand function is quadratic, the data points for national sales are plotted and fitted with a second degree polynomial trend line. Excel finds thatD(q) = 0.00005349q2 0.03440302q + 414.53444491where D(q) is the price, in dollars, at which q thousand drives can be sold.We formatted the trend line equation so that its coefficients are displayed with 8 decimal places. Further precision would result in only very minor changes in computed values. This apparent increase in accuracy would be spurious, since it exceeds the accuracy of our market data.

D, R, C, & P, FocusIT(material continues)Class ProjectCMarketing Focus.xlsm54


MarketingComputerDrivesIn our units, the revenue function R(q) is to give the revenue, in millions of dollars from selling q thousand drives. D(q) gives the price, in dollars per drive, at q thousand drives. Hence, D(q)q1,000 gives the revenue, in dollars, from selling q thousand drives. To express this revenue in millions of dollars, we divide by 1,000,000.R(q) = (D(q)q1,000)/1,000,000 = D(q)q/1,000D, R, C, & P, FocusIT

(material continues)Class ProjectCMarketing Focus.xlsm55D, R, C, & P, Focus


MarketingComputerDrivesIt appears that we might obtain the maximum revenue, if we sold approximately 1,400 thousand drives. Cells C91:E91 in the sheet Functions of Marketing Focus.xlsm allow us to compute D(q) and R(q) for any value of q. We find that D(1,400) = $261.53 and R(1,400) = 366.142 million dollars.We want the total cost function, C(q), to give the cost, in millions of dollars, of producing q thousand drives.C(q) = fixed cost + variable cost for q thousand drives

IT



MarketingComputerDrivesLet marginal cost denote the price, in dollars per drive, for production at a given number of drives. Our cost information can now be summarized in the following table. The Batch Size column displays the numbers of drives, in Ks, that will be produced in first, second, and all further production lots.

C(q) depends upon 7 numbers: q (Quantity), Fixed Cost, Batch Size 1, Batch Size 2, Marginal Cost 1, Marginal Cost 2, and Marginal Cost 3.

IT(material continues)Class ProjectCMarketing Focus.xlsm57D, R, C, & P, Focus


MarketingComputerDrivesFor example, suppose that we produce 1,400 thousand drives. The fixed cost is 135 million dollars. The first 800 thousand drives cost $160 per drive, for a total cost of 8001,000160/1,000,000 = 800160/1,000 = 128 million dollars. The next 400 thousand drives cost $128 per drive, for a total cost of 4001,000128/1,000,000 = 400128/1,000 = 51.2 million dollars. The remaining 200 thousand drives cost $72 per drive, for a total cost of 2001,00072/1,000,000 = 20072/1,000 = 14.4 million dollars.Adding these, we find that C(1,400) = 135.0 + 128.0 + 51.2 + 14.4 = 328.6 million dollars.It is possible, though very tedious, to use nested IF functions in Excel to produce a formula for C(q). This is demonstrated in Cell D125 of the sheet Functions in Marketing Focus.xlsm.IT(material continues)Class ProjectCMarketing Focus.xlsm58D, R, C, & P, Focus


MarketingComputerDrivesWhen it is difficult or very complicated to compute with built-in Excel functions, we can seek help from custom programming. In particular, the Visual Basic language can be used to construct user defined functions in any Excel file. These are accessed via pull-down menus, in the same way that the built-in functions are used.Given the complexity of our cost function, it is desirable to custom program the formula in Visual Basic and attach it to the Excel file. This has been done in Marketing Focus.xlsm, with the resulting function, COST, being placed in the list of functions under User Defined. Look in the formula bar for Cell E125 in the sheet Functions of Marketing Focus.xlsm to see how the function is used.To view the Visual Basic code for COST in Office 2007, click on the Developer tab, then on Visual Basic. In earlier versions, click on Tools/Macro/Visual Basic Editor. In the side bar, select VBAProjects (Marketing Focus.xlsm), double click on Modules, then double click on Module1.IT(material continues)Class ProjectCMarketing Focus.xlsm59D, R, C, & P, Focus


MarketingComputerDrivesYou are not expected to do any Visual Basic programming in Mathematics for Business Decisions. However, you will probably find it convenient to use the function COST while working on your teams project. To transfer the Cost function from Marketing Focus.xlsm into another Excel file, open both files. Open Module1 (as directed above), click on its icon in the side bar under VBAProject (Marketing Focus.xlsm) and drag the icon into the VBAProject side bar area of the new file.You can also move modules containing macros or custom functions with export/import. To do this, open the module, click on File/Export File, then select a folder to contain a copy of the module. To attach the module to another Excel file, open the Visual Basic Editor in that file, click on File/Import File, and open the module. IT(material continues)Class ProjectCMarketing Focus.xlsm60D, R, C, & P, Focus


MarketingComputerDrivesThe Cost function can be added to Graphing.xlsm and plotted with that utility, or it can be plotted as in the sheet Functions of Marketing Focus.xlsm. It is most interesting to display the graphs of both R(q) and C(q) on the same set of axes.We will make a profit when R(q) > C(q) and will operate at a loss where R(q) < C(q). It appears that we can make a profit by selling between approximately 650 and 1,650 thousand drives. Verify this with numerical experimentation in Cells B170:F170 in the sheet Functions of Marketing Focus.xlsm.IT



MarketingComputerDrivesOur main interest is in profit, which is given by revenue minus cost. Let P(q) be the profit, in millions of dollars, from selling q thousand drives.P(q) = R(q) C(q)A plot of the profit function has been copied from Marketing Focus.xlsm into the next Focus page. Each of the two humps represent a local maximum profit. It appears that the local maximum that occurs on the right, at approximately 1,280 thousand drives, is slightly higher than the local maximum on the left. Hence, the largest possible profit, or global maximum, is approximately P(1,280). This is close to 42 million dollars. Looking at the plot of the demand function, we estimate that the 1,200 thousand drives should be priced at approximately $280 each. Verify this with numerical experimentation in Cells B170:F170 in the sheet Functions of Marketing Focus.xlsm.What we need now is some way to replace graphical estimates with more precise computations.IT(material continues)Class ProjectCMarketing Focus.xlsm62D, R, C, & P, Focus


MarketingComputerDrivesWHAT SHOULD YOU DO?Each team should now find formulas for, and make plots of, its demand, revenue, cost, and profit functions. Work with the same units and coefficient precision as in the Class Project. Use your plots to estimate the production range that would yield a profit, and to estimate the maximum possible profit.IT

(material continues)Class ProjectCMarketing Focus.xlsm63Differentiation, Marginal

We can use a calculator or Excel to compute values of MC(q). Forexample,

Thinking in terms of money, the marginal cost at the level of 1,000 dinners, is approximately $8.88 per dinner. Similar computations show thatMC(2,000) $6.88 and MC(3,000) $5.93.Since the marginal cost per dinner depends upon the number of dinners currently being prepared, it is helpful to look at a plot of MC(q) against q. This is created in the sheet M Cost of the Excel file Dinners.xlsm.(material continues)Differentiation. Marginal Analysis: page 2ExcelITC65

Differentiation, MarginalDifferentiation. Marginal Analysis: page 3Dinners.xlsm(material continues)Visiting that file, we see that the First Plan marginal cost in Column D is very large, when only a small number of dinners are prepared. MC(0) = $177.00/dinner. However, the cost per dinner drops rapidly. MC(100) is down to $20.63/dinner, and MC(3,999) = $5.34/dinner.When we talk about the cost per dinner, when q dinners are being prepared, it is not clear whether we should consider the cost of the qth or the (q + 1)st dinner. Rather than arbitrarily choosing to move ahead to the (q + 1)st dinner, as in the First Plan, we get a better indication of the marginal cost at q by averaging the change in both directions from q. We will call this the Second Plan for computing marginal cost.ITC66Differentiation, Marginal

(material continues)Differentiation.Marginal Analysis: page 4The marginal costs, computed with the Second Plan, are shown in Column E of M Cost in the Excel file Dinners.xlsm.What can we learn from MC(q)? The restaurant managers might want to know how many dinners would need to be prepared per week in order to get the price per dinner for further dinners down to $8 or less. Glancing down Column E in the sheet M Cost shows that MC(q) $8 for q 1,327.One further refinement is necessary in our computation of marginal cost. MC(q) is best defined as the instantaneous rate of change in total cost,per unit. The formula is only an average change over aninterval of length 2.

ITDinners.xlsmC67

To improve the computation, we let mathematics do something that cannot be done in the real world. We consider a non-integer change in the number of dinners that are prepared. If h is any small number then

is the average change over an interval of length 2h. Since the numerator is a difference of two numbers and the entire expression is a quotient, it is usually called a difference quotient. As h is assigned smaller and smaller values, the difference quotient gives better and better approximations for MC(q).

Differentiation, Marginal(material continues)Differentiation. Marginal Analysis: page 5We indicate that MC(q) is an instantaneous rate of change, by writingITDinners.xlsmC

68Differentiation, MarginalIn practice, we will compute marginal cost by evaluating the differencequotient at a small value of h. Since excessively smallvalues may cause numerical problems, we will often use h = 0.00001.This method is the Final Plan, that is used in Column F of M Cost in the Excel file Dinners.xlsm.

(material continues)Differentiation. Marginal Analysis: page 6 Explain, in terms of real world dinners and dollars, why it is plausible that, for buffalo steak dinners, MC(q) is very large for small values of q, and that it is always positive.Exercise 1

Use Excel, with h = 0.00001, to compute MC(1,000) for buffalo dinners, rounded to 2 decimal places. Exercise 2ExcelITDinners.xlsmC69Differentiation, MarginalMarginal analysis is also used with revenue, demand, and profit functions. For example, the marginal revenue, MR(q), at q dinners is the instantaneous rate of change in revenue per dinner, when q dinners are being prepared and ordered at the demand price per dinner. This is defined by

and is computed by evaluating the difference quotient for asmall value of h. Similar definitions apply to the marginal demand, MD(q), and marginal profit, MP(q), which are defined by the following limits.

(material continues)Differentiation. Marginal Analysis: page 7I Use Excel, with h = 0.00001, to compute MC(2,500) for buffalo dinners, rounded to 4 decimal places. Exercise 3TDinners.xlsmC70Differentiation, MarginalMany aspects of the demand function are reflected in properties of the difference quotients for marginal demand, and in the marginal demand function. D(q) is always decreasing. Hence, all difference quotients for marginal demand are negative, and MD(q) is always negative. The more rapidly D(q) drops, the more negative are the difference quotients, and the further negative is MD(q).

Differentiation. Marginal Analysis: page 8(material continues)IValues for all of our mar-ginal functions are computed in the sheets M Cost and M Profit of the Excel file Dinners.xlsm. The graphs of MD(q), MR(q), and MP(q) are also displayed in those sheets.TDinners.xlsmC71Differentiation, MarginalMarginal analysis can tell us a great deal about the profit function. Refer back to these plots while reading the next pages.

Differentiation. Marginal Analysis: page 10(material continues)ITDinners.xlsmC73Differentiation, MarginalSince P(q) = R(q) C(q), profit will increase with more dinners if the increase in revenue per dinner is greater than the increase in cost per dinner. This happens where MR(q) > MC(q). Similarly, profit will decrease with more dinners if the change in revenue per dinner (positive or negative) is less than the increase in cost per dinner. This happens where MR(q) < MC(q).Profit stops increasing, and starts to decrease at its maximum value. Hence, the maximum profit must occur where MR(q) = MC(q). From the plot of MR(q) and MC(q), it appears that the two graphs cross at a point where q is slightly greater than 2,000. The computations in the sheet M Profit of Dinners.xlsm show that MR(2,025) = $6.84 = MC(2,025). Hence, the maximum profit will occur at q = 2,025 dinners. Direct computation shows that D(2,025) = $22.21 and that P(2,025) = $14,052.Considering only the mathematics of our analysis, the restaurant chain should expect to sell 2,025 buffalo steak dinners per week, priced at $22.21 per dinner, for a total profit of $14,052. Note that all of this information is consistent with the estimates that we made in the section Demand, Revenue, Cost, and Profit.(material continues)Differentiation. Marginal Analysis: page 11ITDinners.xlsmC74

Differentiation, MarginalThis must be where MP(q) changes from positive to negative. Thus, the maximum profit occurs when the marginal profit is zero, MP(q) = 0.The computations in Column F of M Profit in Dinners.xlsm show that MP(2,025) = $0.00. This agrees with the value found from the marginal analysis of revenue and cost.Marginal analysis of profit offers another way to determine the maximum profit. Where profit, P(q), is increasing, marginal profit, MP(q), is positive. Where P(q) is decreasing, MP(q) is negative. The change from increasing to decreasing profit occurs at the maximum profit.(material continues)Differentiation. Marginal Analysis: page 12ITDinners.xlsmC75Differentiation, MarginalExercises 4, 5, and 6 refer to the situation that was discussed in Exercises 3 and 14 of Demand, Revenue, Cost, and Profit. The demand and cost functions for a good were given by D(q) = 0.1q + 150 and respectively.

Differentiation. Marginal Analysis: page 13 Use the methods of Marginal Analysis to plot the marginal cost function, MC(q), and the marginal revenue function, MR(q), on the same set of axes.Exercise 4 (i) Use the methods of Marginal Analysis to plot MP(q). (ii) Use your graph to estimate the value of q that maximizes profit. (iii) What is the maximum possible profit? (iv) At what price should the good be sold, in order to realize the maximum profit?Exercise 5 (i) Use your graphs from Exercise 4 to estimate the value of q that makes MR(q) = MC(q). (ii) How does this value relate to your work in Exercise 5?Exercise 6(material continues)ExcelITC76Differentiation, MarginalExercises 7, 8, and 9 refer to the situation that was discussed in Exercises 4 and 15 of Demand, Revenue, Cost, and Profit. The demand and cost functions for one model of audio speaker were given by D(q) = 0.00006q2 + 250 and C(q) = 60,000 + 110q, respectively.Differentiation. Marginal Analysis: page 14 Use the methods of Marginal Analysis to plot the marginal cost function, MC(q), and the marginal revenue function, MR(q), on the same set of axes.Exercise 7 (i) Use the methods of Marginal Analysis to plot MP(q). (ii) Use your graph to estimate the value of q that maximizes profit. (iii) What is the maximum possible profit? (iv) At what price should the speakers be sold, in order to realize the maximum profit?Exercise 8 (i) Use your graphs from Exercise 7 to estimate the value of q that makes MR(q) = MC(q). (ii) How does this value relate to your work in Exercise 8?Exercise 9(material continues)ExcelITC77Differentiation, MarginalExercises 10 and 11 refer to the demand and cost functions for car alarm systems that are plotted in Exercises 5 and 16 of Demand, Revenue, Cost, and Profit.Differentiation. Marginal Analysis: page 15 Use your conceptual understanding of marginal demand and the plot of the demand function to sketch a rough graph of MD(q).Exercise 10 Use your conceptual understanding of marginal cost and the plot of the cost function to sketch a rough graph of MC(q).Exercise 11 Sketch what you believe might be a typical plot of a marginal revenue function. Explain the reasoning behind your sketch.Exercise 12(material continues)ExcelITC78Differentiation, DerivativesDifferentiation. Derivatives(material continues)2. DERIVATIVESDifference quotients are used in many business situations, other than in marginal analysis. For this reason, it is convenient to have a standard name and common notation for limits of such quotients.Let f be a function which is defined on an open interval containing areal number x. Suppose that approaches a number m as his taken to be smaller and smaller. We write

and call m the derivative of f at x.To show that m depends on both the function, f, and the number, x, we denote the derivative by f (x).

ITC79Differentiation, DerivativesExample 3. Let f(x) = x3 + 6. Use Excel to plot both f(x) and f (x) over the interval from 3 to 3. The work is shown in the file Example 3.xlsx and the resulting graph is shown on the following page.

Example 3.xlsxDifferentiation. Derivatives: page 3(material continues)The following computation can be done with a calculator or with Excel. If it is likely that we will need to evaluate f (x) for several different values of x, then Excel is a more efficient tool.ITC81

Let f(x) = 2x. (i) Use Excel and the method of Example 3 to plot the graph of f over the interval [0, 4]. (ii) Use the interpretation of the derivative as an instantaneous rate of change to explain the form of your graph.Exercise 14Differentiation, DerivativesDifferentiation. Derivatives: page 4 Let f(x) = 1 ex/2. Use Excel and the method of Example 3 to plot both f(x) and f (x) over the interval from 0 to 10.Exercise 13(material continues)Graphing.xlsmITExample 3.xlsxC82Differentiation, DerivativesDifferentiating.xlsm(material continues)IDifferentiation. Derivatives: page 5The evaluation of difference quotients provides a good understanding of differentiation, but it is rather tedious. Carrying out this much work, while in the middle of a business application, could easily distract us from the basic business problem.What we are doing is called numerical differentiation. Many software packages, but (unfortunately) not Excel, have built-in functions that perform numerical differentiation. To meet this need, we have created an Excel file Differentiating.xlsm, that serves as a numerical differentiation utility. To use Differentiating.xlsm, open the file and follow the instructions in the blue box. In particular, be sure that Excel is set for Automatic Calculation. The file works in much the same way as Graphing.xlsm. Once you have entered the formula for f(x), you can get numerical values for f(x) and f (x), and can see the graphs of these functions over any interval.TC83Differentiation, DerivativesFrom now on, unless otherwise indicated, we will assume that all differentiation is performed with Differentiating.xlsm.When entering a function in Differentiating.xlsm, you can use the letters s, t, u, v, or w as constants. If this is done, you must enter values in Cells O23:O27 for any letters that you use.For example, suppose that you want to differentiate f(x) = x s, where s is a number that is in Cell O23. You would enter = x^s in Cell C24.Accept the function and then click on the Differentiate button at the right of the plots. Since Excel is set for automatic calculation, entering a new number in Cell O23 will automatically change the function, redraw the plots, and recompute the derivative.(material continues)IDifferentiation. Derivatives: page 6Click here for information on running the macro in Differentiating.xlsm.TDifferentiating.xlsmC84Differentiation, DerivativesDifferentiation. Derivatives: page 7(material continues)IIf values are entered in Cells O23:O27; then constants s, t, u, v, or w may also be used to describe the range, [a, b], of the plot. If these constant names are used, they must be entered in Cells I24:J24 preceded with "=" signs. For example, =s. If f '(x) is constant, the displayed plot in Differentiating.xlsm will be distorted. To correct this, format the y-axis to have realistic fixed minimum and maximum values.Caution!Example 4. Use Differentiating.xlsm to redo Examples 1 and 2. Open Differentiating.xlsm and enter =x^3+6 in Cell C24. Click on to accept the function, and then click on the Enter button on the Differentiating toolbar. Enter 2, 3, and 3 in Cells E24, I24, and J24, respectively. The resulting computation and plot of both the function and its derivative are shown on the following page.TDifferentiating.xlsmC85

Differentiation, DerivativesDifferentiation.Derivatives: page 8(material continues)I Use Differentiating.xlsm to redo Part i of Exercise 5.Exercise 15 Use Differentiating.xlsm to redo Part i of Exercise 8.Exercise 16TDifferentiating.xlsmC86Differentiation, DerivativesExercises 17, 18, and 19 refer to the situation that was discussed in Exercises 3 and 14 of Demand, Revenue, Cost, and Profit. The demand and cost functions for a good were given by D(q) = 0.1q + 150 and respectively.

Use Differentiating.xlsm to plot marginal cost, C, over the interval from 0 to 1,500.Exercise 17 (i) Use Differentiating.xlsm to plot marginal profit, P over the interval from 0 to 1,500. (ii) Experiment with the Computation boxes to find a value for q that is greater than 100, and at which P(q) = 0. (Hint: This value of q might not be an integer.)Exercise 18Differentiation. Derivatives: page 9 (i) Use Differentiating.xlsm to plot marginal revenue, R over the interval from 0 to 1,500. (ii) Experiment with the Computation boxes to find a value for q at which R(q) = 0.Exercise 19(material continues)ITDifferentiating.xlsmC87Differentiation, Derivatives(material continues)Exercises 20, 21, and 22 refer to the situation that was discussed in Exercises 4 and 15 of Demand, Revenue, Cost, and Profit. The demand and cost functions for audio speakers were given by D(q) = 0.00006q2 + 250 and C(q) = 60,000 + 110q, respectively. Use Differentiating.xlsm to plot marginal revenue, R, over the interval from 0 to 2,100.Exercise 20 (i) Use Differentiating.xlsm to plot marginal profit, P, over the interval from 0 to 2,100. (ii) Experiment with the Computation boxes to find a value for q at which P(q) = 0. (Hint: This value of q might not be an integer.)Exercise 21I Use Differentiating.xlsm to plot marginal cost, C, over the interval from 0 to 2,100. Hint: remember the caution about using Differentiating.xlsm, when the graph of f is distorted.Exercise 22Differentiation. Derivatives: page 10TDifferentiating.xlsmC88Differentiation, Properties & ApplicationsDifferentiation. Properties and Applications(material continues)3. PROPERTIES AND APPLICATIONSA few basic properties of derivatives can save us time while working on business problems. Since a derivative represents the instantaneous rate of change in a function, it is not surprising that multiplying a function, f, by a constant, a, also multiplies its derivative by the same constant.If g(x) = af(x), then g(x) = af (x).Thus, for example, doubling the production costs for a product will double the marginal cost.The derivative of the sum or difference of two functions is the sum or difference of their derivatives.If h(x) = f(x) g(x), then h (x) = f (x) g(x).This property is often used with marginal profit. If R, C, and P, are the revenue, cost, and profit functions for a good, then P(q) = R(q) C(q). Thus, P(q) = R(q) C(q).ITC89Differentiation, P. & A.Differentiation. Properties & Applications: page 2(material continues)Since the marginal profit is equal to the marginal revenue minus the marginal cost, the marginal profit is equal to zero exactly when the marginal revenue is equal to the marginal cost.Some useful properties of differentiation are clear from the definition of a derivative. A function, f, is constant on an interval, if f(x) is the same number for all values of x in the interval. For example, we might have f(x) = 100 for all numbers, x, with 0 x 10. Since the function does not change, the derivative of a constant function is zero.Approximately constant demand functions are encountered in the case of some commodities, such as wheat or corn, where a change in production level by a single farmer has no effect on the price per bushel for his or her crop. In such a case the marginal demand is zero.A function of the form f(x) = mx + b, where m and b are constants, is called linear. We know that the derivative of f is the sum of the derivatives of mx and b. Since b is a constant, its derivative is zero. The graph of y = mx is a straight line, with a slope of m. Thus, the instantaneous rate of change for mx is always equal to m.ITC90Differentiation, P. & A.Differentiation. Properties & Applications: page 3(material continues)Putting these facts together, we see that, if f(x) = mx + b, then f (x) = m, for all values of x.Linear functions are often used as models for parts of demand and cost functions. In this case, there is no need for numerical differentiation. For example, if the cost function for a good is given by C(q) = 120q + 5,000, then its marginal cost is C(q) = 120, for all quantities, q. If f(x) = 0.75x + 1.94, find a formula for f (x).Exercise 23 Let f(x) = 3g(x) + 12 and suppose that g(5) = 2. What is f (5)?Exercise 24 When 12,000 mountain bicycles are being produced and sold the marginal revenue is $895 and the marginal cost is $787. (i) What is the marginal profit at this production level? (ii) What does this number mean in terms of bicycles and dollars?Exercise 25ITC91Differentiation, P. & A.Differentiation. Properties & Applications: page 6(material continues)Suppose that C(t) is the total yearly cost if drives are ordered every t days, starting from the first of the year. Our goal is to find the value of t that minimizes C(t).If orders are placed every t days, then there will be 365/t orders peryear, and each order will be for drives. In order to

understand the problem, let It(x) be the number of drives in the companys inventory at x days after the arrival of an order. As an example, we will plot It(x), assuming that the company orders every 30 days, that is for t = 30. Inthis case, each order will be for drives.

Note that, on average, there will be one-half the number of parts that are ordered in storage.ITC94In general, if the company orders every t days, the mean number ofdrives in the inventory will be

Differentiation, P. & A.(material continues)Differentiation. Properties & Applications: page 7I

TC95

Differentiation, P. & A.At one cent per day per drive for storage, it will cost $0.01365 = $3.65 per year to store a single drive. We can now set up a formula for C(t).Differentiation. Properties & Applications: page 8

(material continues)ITDifferentiating.xlsmC96Differentiation, P. & A.(material continues)Differentiation. Properties & Applications: page 10Where f(x) is decreasing, its difference quotients are negative, and it derivative is negative. Where f(x) is increasing, f (x) is positive. Hence, the minimum value of f(x) occurs where f (x) = 0. This is exactly what is shown in the graphs.By experimenting with the values of x in Cell E24 of Differentiating .xlsm, we find that f (41.8569) = 0.000 and that f(41.8569) = 1,020,928. Computation shows that 41.8569 days is equal to 41 days, 20 hours, 33 minutes, and 56 seconds. While this is very precise, it is not a highly practical time interval at which to place orders! The natural plan would be to round this time to 42 days. Putting 42 into Cell E24 of Differentiating.xlsm, shows that f(42) = 1,020,929. Since this is only $1 more than the minimum possible value of f(x), it does no harm to order every 42 days.Returning to the language of our problem, we have found that the most economical plan is to order (50,000/365)t 5,753 drives every 42 days. This will result in a total cost of approximately $1,021,000 per year for procuring and storing the drives.ITDifferentiating.xlsmC98Differentiation, P. & A. Redo Example 5, assuming that the fixed cost of an order is $1,000 and that it costs 1.5 cents per day to store a CD drive.Exercise 29 Redo Example 5, assuming that the fixed cost of an order is $1,500 and that it costs 0.8 cents per day to store a CD drive.Exercise 30Differentiation. Properties & Applications: page 11 (i) Use your teams data to plot MR, MC, and MP for its product. (ii) Prepare computational cells, as in the sheet Marginal of Marketing Focus.xlsm and use these to answer your teams Questions 1-4 for its product. Use the same units as in the work on the Class Project in Marketing Focus.xlsm. Study of the Focus pages will help with this work.Exercise 31(material continues)ITDifferentiating.xlsmC99* This section is not needed for the study of other material in Mathematics for Business Decisions.Differentiation, Tangents & SlopesDifferentiation. Tangents & Slopes(materialcontinues)4. TANGENTS & SLOPESThe subject of differentiation is much more sophisticated than we have indicated in our presentation. However, the numerical methods that we have introduced provide good intuition, and are adequate for a wide range of simple business problems. Differentiation is one of the main components of the mathematical subject, calculus.There is a geometric interpretation of the derivative that often helps in the understanding of business graphs. Recall that the slope of a straight line is defined to be the change in height divided by the horizontal change between any two points on the line. We would like to extend this notion to define slope for the graph of a function f at a single point (x, f(x)).Example 6. Let and consider the point (1, f(1)) = (1, 1). Moving an increment h units to the left and to the right of (1, 1), locates the points (1 h, f(1 h)) and (1 + h, f(1 + h)) on the graph of f. The straight line connecting these points is the secant line determined by the increment h.

ITC100Differentiation, T. & S.Differentiation. Tangents & Slopes: page 2Notice that the slope, mh, of the blue secant line is given by the difference quotient

The red line that we will call tangent to the graph of f at the point (1, f(1)) has a slope which is the limit of the slopes of the secant lines, as the increment, h, approaches 0. That is,

I

2hf(1 + h) f(1 h)(1, f(1))y(material continues)TC101

Since we see that the derivative, f (1), is the slope of the tangent line at (1, f(1)). The Excel file Slope.xlsx plots both the tangent and secant lines for selected values of h. Open that file and experiment with different increments. For relatively small values of h, the blue secant line appears to coincide with the red tangent line.What we have just learned about the tangent line for at the point (1, f(1)) is also true for any differentiable function. If f(a) exists, then the slope of the tangent line to the graph of f at the point (a, f(a)) is defined to be f(a). For this reason, f(a) is called the slope of the graph of f at the point (a, f(a)). This is often shortened to the slope of f at a.In practical situations, the slope, f (a), of f at (a, f(a)) is interpreted as the instantaneous rate of change of f(x) at a. This follows from the fact that the slope mh of a secant line is the average rate of change in f(x) over the interval from a h to a + h. The limit of this average rate, as the increment, h, approaches 0, is the instantaneous rate of change.Differentiation, T. & S.(material continues)Slope.xlsx

Differentiation. Tangents & Slopes: page 3ITC102Differentiation, T. & S.(material continues)Since we know the slope of the tangent line and have one point on that line, we can determine its equation. If f (a) exists, then the tangent line to the graph of f at the point (a, f(a)) has the equationy = f (a)(x a) + f(a).Differentiation. Tangents & Slopes: page 4 When using the equation for a tangent line it is important to distinguish between a and x. a is the first coordinate of the point at which the line is tangent. x is simply the independent variable in the equation of the tangent line.CautionExample 7. Find the slope of the graph of f(x) = x2 4x + 4 at the point (3, f(3)), and find an equation for the tangent line at that point.Differentiating.xlsm shows that f (3) = 2. Hence, the slope of f is 2 at x = 3. This means that f(x) is increasing at an instantaneous rate of 2 units per unit increase in x, when x = 3. In the tangent line equation, we have a = 3, and f(a) = f(3) = 32 43 + 4 = 1.ITDifferentiating.xlsmC103 (i) Find the slope of the graph of f(x) = 2x at the point (1, f(1)), and (ii) find an equation for the tangent line at that point. (iii) Use Excel to show both the graph of f and the graph of the tangent line in a single plot. Plot both functions over the interval [0, 3].Exercise 32Differentiation, T. & S.(material continues)Tangent Line

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yDifferentiation. Tangents & Slopes: page 5TDifferentiating.xlsmC104 (i) Find the slope of the graph of the natural logarithm function, f(x) = ln(x) at the point (2, f(2)), and (ii) find an equation for the tangent line at that point. (iii) Use Excel to show both the graph of f and the graph of the tangent line in a single plot. Plot both functions over the interval [1, 5].Exercise 33 Let f(x) = 1 + x/2. (i) Find the slope of the graph of f at the point (4, f(4)), and (ii) find an equation for the tangent line at that point. (iii) Use Excel to show both the graph of f and the graph of the tangent line in a single plot over the interval [0, 6]. (iv) Use the formulas for differentiation to explain what happened in Part iii.Exercise 34 Many formulas have been developed which allow us to find the derivative of a function symbolically, without the need for numerical approximation. These will not be needed in our work with differentiation, but they might be of interest to you. To see a few examples, click on symbolic.InformationDifferentiation, T. & S.(material continues)Differentiation. Tangents & Slopes: page 6ITExcelDifferentiating.xlsmC105Differentiation, FocusHow can differentiationhelp us price 12-GB drives?We must consider the effect ofthe units that we are using. Recall; from the sheet Functions in Marketing Focus.xlsm and from the Focus pages of the section Demand, Revenue, Cost, and Profit; that C(q) gives the cost, in millions of dollars for producing q thousand drives. C'(q) gives the rate of change in C(q), with respect to q.


MarketingComputerDrivesIMarketing Focus.xlsm(materialcontinues)Class ProjectT

C106Differentiation, Focus


MarketingComputerDrivesHence, C'(q) gives marginal cost in millions of dollars per thousand drives, at a production level of q thousand drives.From our variable cost data, we see that the marginal cost for q thousand drives is $162 per drive if q is less than or equal to 800; $128 per drive if q is greater than 800, but less than or equal to 1,200; and $72 per drive for q greater than 1,200. These values give marginal cost in dollars per drive at a production level of q thousand drives.How does this relate to the derivative? 1,000,000C'(q) is in dollars per thousand drives, and (1,000,000C'(q))/1,000 is in dollars per drive. The last quantity simplifies to 1,000C'(q). Hence, where C is differentiable, 1,000C'(q) gives the marginal cost, MC(q), in dollars per drive, at a production level of q thousand drives.ITMarketing Focus.xlsm(materialcontinues)Class ProjectC107Differentiation, Focus


MarketingComputerDrivesFor 12-GB drives, R(q) gives the revenue, in millions of dollars, from the sale of q thousand drives, and R'(q) gives the rate of change in R(q), with respect to q. Hence, R'(q) gives marginal revenue in millions of dollars per thousand drives, at a production level of q thousand drives.Since our data for marginal profit occurs naturally in dollars per drive, we will us these units for all 12-GB drive marginal functions. (1,000,000R'(q))/1,000 is in dollars per drive. This simplifies to 1,000R'(q). MR(q) = 1,000R'(q) is the marginal revenue, MR(q), in dollars per drive, when q thousand drives are being sold.The functions MR(q), and MC(q) are plotted in the sheet Marginal of Marketing Focus.xlsm. These graphs are also shown on the next page. Note that the marginal revenue is positive where the revenue function is increasing and negative where revenue is decreasing. Since cost increases with quantity, marginal cost is always positive.ITMarketing Focus.xlsm(materialcontinues)Class ProjectC108P(q) = R(q) C(q) gives the profit, in millions of dollars, from the sale of q thousand drives. Since P(q) = R(q) C(q), we know that, in dollars per drive, MP(q) = MR(q) MC(q) = 1,000R'(q) 1,000C'(q) = 1,000(R'(q) C'(q)) = 1,000P'(q).

on the projectCalculus, Mathematics, Tests, Homework, Computers MarketingComputerDrives

Differentiation, FocusITMarketing Focus.xlsm(materialcontinues)Class ProjectC109

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Calculus, Mathematics, Tests, Homework, Computers MarketingComputerDrivesSince, in dollars per drive, MP(q) = 1,000P'(q), we see that MP(q) and P'(q) always have the same sign, and are equal to zero at exactly the same values of q. Hence, MP(q) = 0 if and only if R (q) = C (q).The above graphs are copied from the sheet Marginal in Marketing Focus.xlsm. The changes in cost per drive at q = 800 and q = 1,200 thousand drives cause drops in the plot of MC(q) and produce corresponding jumps upward in MP(q). Due to the second gap, marginal profit crosses the axis in two places. These reflect the two local maxima, or humps, in the profit function.Differentiation, FocusIT

Marketing Focus.xlsm(materialcontinues)Class ProjectC110Differentiation, Focus


MarketingComputerDrivesFrom the plot of marginal profit, it appears that MP(q) = 0 when q is approximately equal to 1,150 or 1,250. Looking at the plot of profit, it seems likely that profit has local maxima at the zeros of MP.Cells B107:I107 of the sheet Marginal in Marketing Focus.xlsm compute values for all of our functions, when a value of q is entered in B107. Experimentation with different values of q produces the following results.

As we have guessed from the graphs, the local maximum at the larger value of q is greater than at the smaller value of q.According to our model, Card Tech would earn a maximum profit of $42,176,000 from selling 1,262,270 12-GB drives, priced at $285.88. This answers Questions 1, 2, and 3 from the original discussion of our Class Project.

ITMarketing Focus.xlsm(materialcontinues)Class ProjectC111Differentiation, Focus


MarketingComputerDrivesWe will answer Question 4 for our Class Project, by investigating the sensitivity of the maximum profit to changes from the optimal price. Numerical experimentation in Cells B107:I107 of the sheet Marginal in Marketing Focus.xlsm shows that raising the price per drive by $2 to $287.88 would lower the number of drives sold to 1,250,431, and decrease the profit to 42.143 million dollars. This is a drop of only 0.08% from the maximum possible profit.Similar work shows a 0.08% drop in profit if the price per drive is lowered to $283.88. Clearly, profit is not very sensitive to $2 changes in price. On the other hand, a $10 change in price would cause a noticeable effect on profits. Raising the price per drive to $295.88, drops the profit by 2.01% and lowering the price to $275.88, drops the profit by 1.91%.ITMarketing Focus.xlsm(materialcontinues)Class ProjectC112Differentiation, Focus


MarketingComputerDrivesSince profit is not very sensitive to small changes in price, Card Tech would probably adjust the optimum price of $285.88 to a more common value, such as $284.99 or $285.99. Experimentation in the sheet Marginal of Marketing Focus.xlsm, shows that these could be expected to reduce the profit by only 0.02%.

WHAT SHOULD YOU DO?Each team should now plot MR, MC, and MP for its product and prepare computational cells, as in the sheet Marketing. Use these to answer your teams Questions 1-4 for its product. Use the same units as we have used in the Class Project.ITMarketing Focus.xlsm(materialcontinues)Class ProjectC113

To use Solver, click on Data, then Solver in the Analysis box. In older versions of Excel select Tools in the main Excel menu, then click on Solver.Using Solver, Solver

WHLFRAGILECrush slowlyUsing Solver. Excels Solver: page 2(material continues)Computer Problem?ITTo use Solver, click on Data, then Solver in the Analysis box. In older versions of Excel select Tools in the main Excel menu, then click on Solver.Enter cell that computes volume.Select Max.Enter cells that contain dimensionsClick on Add.Shipping.xlsxC115

Using Solver, SolverUsing Solver. Excels Solver: page 3(material continues)ITEnter cell that computes circumference.Select = 0 and D7 >= 0.TShipping.xlsxC

118 Big Box shipping company limits the circumference of its boxes to at most 80 inches with the sum of their dimensions to be at most 150 inches. (i) Modify Shipping.xlsx and use Solver to find the dimensions of the box with maximum volume that will be accepted by Big Box. (ii) What is the maximum volume of a box which Big Box will ship?Exercise 2Using Solver, SolverUsing Solver. Excels Solver: page 6(material continues) Rush! shipping company limits the size of the boxes that it accepts by limiting their volume to at most 16 cubic feet (27,648 cubic inches). For it to ship a box, each dimension must be between 3 and 54 inches. (i) Modify Shipping.xlsx and use Solver to find the dimensions of a Rush! box which will accept the longest possible item. Hint: Use different initial values for each dimension. (ii) What is the maximum length of such an item? Note that the longest item which can be shipped in a box has a length of

ITExercise 3Shipping.xlsxC119Using Solver, SolverExample 2. Return to the restaurant management example that has been studied in Demand, Revenue, Cost and Profit; and Differentiation. We will consider the demand, profit, and marginal profit functions; D(q), P(q), and MP(q); that were developed in those sections. Recall that a restaurant chain is planning to introduce a new buffalo steak dinner. D(q) is the price, in dollars, at which q dinners per week would be ordered. P(q) is the profit from selling q dinners priced at D(q) dollars and MP(q) =P(q) is the marginal profit.What is the minimum number of dinners that the restaurants would have to sell in order to make a profit?In the sheet Solver of Dinners.xlsm we have plotted D(q), P(q), andP(q); and have built computation boxes that evaluate the functions at q. Referring to the graph in that sheet we see that P(q) is first positive when q is approximately equal to 700. Solver is now used to find a value close to 700, in Cell G11 that makes Cell I11 equal to 0.Using Solver. Excels Solver: page 7(material continues)IDinners.xlsmTC120

Using Solver, SolverUsing Solver. Excels Solver: page 8(material continues)ITCDinners.xlsm

121Using Solver, SolverUsing Solver. Excels Solver: page 9(material continues)Solver finds that 689.51027 dinners would be needed to produce a non-negative profit. Since this makes no sense in terms of actual meals, the restaurants would need to serve at least 690 buffalo steak dinners in order to make a profit. Recomputingthe Excel sheet shows that thiswould still require a price of atleast $29.13 per dinner.Three questions are of great interest to the restaurant managers. (i) How should the buffalo steak dinners be priced in order to obtain the maximum profit? (ii) How many dinners per week could they expect to sell at that price? (iii) What maximum profit can they expect?We can use Solver and the computation boxes in the sheet Solver of Dinners.xlsm to answer all of these questions. We start with a guess value of 2,100 for q, and then maximize P(q).I Different initial values for q may cause Solver to produce slightly different values in the target cell.Warning!TCDinners.xlsm

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Using Solver, SolverUsing Solver. Excels Solver: page 10(material continues)IT

CDinners.xlsm123

Using Solver, SolverUsing Solver. Excels Solver: page 11(material continues)P(q) has a maximum value when q = 2024.8609. As expected, we note that the marginal profit, MP(q), is equal to zero when P(q) is at a maximum. Customers do not order 0.8609 dinners! Hence, q = 2024.8609 is not a practical answer. If we add the constraint that Cell G11 is to be an integer, then Solver can find the whole number of dinners which will maximize profit.ITCDinners.xlsm

124Using Solver, Solver(material continues)Using Solver. Excels Solver: page 12I

If Solver can maximize or minimize any function, why should I use derivatives for optimization?Ask a tutormuch easier way to obtain the same information.Although such precision is clearly unnecessary, we could use Solver to find the expected profit from more realistic pricing. 2,059 dinners, priced at $21.95 would yield an expected weekly profit of $14,040.EntreBuffalo Steak Dinner . . . only $21.95New!This completes our work with the restaurant example. When we first considered the problem in Demand, Revenue, Cost, and Profit; we used graphs to estimate that the restaurant chain should prepare around 2,000 buffalo steak dinners per week and price them at approximately $22, expecting a weekly profit of around $14,000. In Differentiation, we used marginal analysis to refine these estimates, concluding that they should expect to sell 2,025 buffalo steak dinners per week, priced at $22.21, for a total profit of $14,052. Finally, Solver has given us aTCDinners.xlsm125 Use Solver to find the expected maximum profit if the buffalo dinners are priced at (i) $19.95 or (ii) at $24.95.Exercise 5 Use Solver to find the maximum number of dinners that the restaurant chain in our example could sell and still make a non-negative profit.Exercise 4Using Solver, Solver

Ha, ha, ha!!!Using Solver. Excels Solver: page 13(material continues) (i) Use Solver to determine the optimum number of buffalo dinners for the restaurant chain, by finding where MP(q) = 0. (ii) Why do you think that Solver returns a non-integer value for q, even if a constraint is added, requesting only integer values?Exercise 6

Oh, Oh.ITCDinners.xlsm126 (i) Use Solver to find the number qrev of buffalo dinners that will produce the maximum revenue in the restaurant example. (ii) What price should be put on the dinners in order to sell qrev of them? (iii) What maximum revenue can be expected?Exercise 7Using Solver, SolverUsing Solver. Excels Solver: page 14 Consider the situation that was discussed in Demand, Revenue, Cost, and Profit. The demand and cost functions for agood were given by D(q) = 0.1q + 150 andrespectively. (i) Use Differentiating.xlsm to plot profit, P, and marginal profit, P, over the interval from 0 to 1,500. (ii) Use Solver to find a value for q that is greater than 100, and maximizes profit. (iii) Use Solver to find a value for q that is greater than 100, and at which P(q) = 0.Exercise 8

Refer to the good whose demand function is given in Exercise 8. (i) Use Differentiating.xlsm to plot revenue, R, and marginal revenue, R, over the interval from 0 to 1,500. (ii) Use Solver to find a value for q that maximizes revenue. (iii) Use Solver to find a value for q at which R(q) = 0.Exercise 9ExcelIDifferentiating.xlsm(material continues)TCDinners.xlsm127Using Solver, Solver Consider the situation that was discussed in Demand, Revenue, Cost, and Profit. The demand and cost functions for a certain model of audio speaker are D(q) = 0.00006q2 + 250 and C(q) = 60,000 + 110q, respectively. (i) Use Differentiating.xlsm to plot profit, P, and marginal profit, P. (ii) Use Solver to find a value for q that maximizes profit. (iii) Use Solver to find a value for q at which P(q) = 0.Exercise 10 Refer to the audio speakers whose demand function is given in Exercise 10. (i) Use Differentiating.xlsm to plot revenue and marginal revenue. (ii) Use Solver to find a value for q that maximizes revenue. (iii) Use Solver to find a value for q at which R(q) = 0.Exercise 11 In Differentiation, we considered the function C(t) that gives the total yearly expense of purchasing and storing CD drives, if drives are ordered every t days, starting from the first of the year.

Use Solver to find the value of t that minimizes C(t).Exercise 12

(material continues)IUsing Solver. Excels Solver: page 15TExcelDifferentiating.xlsmC128Using Solver, SolverUsing Solver. Excels Solver: page 16(material continues) (i) Use Solver and your teams data to check your earlier numerical results. (ii) Modify your project files and use Solver to answer your teams Questions 6-9. Study of the Focus pages will help with this work.Exercise 14 Consider the Class Project on pricing 12-GB computer drives. (i) Use Solver to find the number q0 of drives, in thousands, such that MR(q0) = 0. (ii) What price should be put on the drives in order to sell q0 drives? (ii) What profit would result from selling q0 drives?Exercise 13Marketing Focus.xlsmIT

C129Using Solver, FocusHow can Solver help usprice 12-GB drives?We will apply Solver to the computation table in Rows 106 and 107 of the sheet Marginal in the Excel file Marketing Focus.xlsm. Starting with initial values of q = 600 and 1,700, and finding values of q that make P(q) = 0, we see that a positive profit will result from selling between 641.462 thousand and 1,666.152 thousand drives.


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Marketing Focus.xlsmC130Using Solver, Focus


MarketingComputerDrivesWith an initial value of 1,300 for q, Solver finds a maximum value of $42,176,000 for P(q). Changing the cell formatting to display three decimal places shows that maximum profit results from the production and sale of 1,262,120 drives. Since D(1,262.120) = $285.91, the theoretical optimum price for the drives is $285.91.It is interesting to note that the lack of sensitivity of profit to price results in rather unstable results when Solver maximizes. The prices and quantities, but not the maximum profit, often depend upon the initial value of q.Marginal analysis provides more accurate results. The computation table in the sheet Marginal shows that MP(1,262.120) = $0.07 per drive. Since there is a positive marginal profit at 1,262.120 thousand drives, the maximum profit will actually occur at a slightly higher production level.IT(materialcontinues)Class ProjectMarketing Focus.xlsmC131Using Solver, Focus


MarketingComputerDrivesTaking an initial value of 1,300 for q and using Solver to find q such that MP(q) = 0, we find a sales level of 1,262.274 thousand drives. These should be priced at $285.88, and their sale will produce a profit of 42.176 million dollars. We will use 1,262,274 drives as our optimum marketing level. In order to duplicate these results, you may need to click on Options in the Solver window, and reset Precision to 0.000001.The values obtained with Solver are in very close agreement with the results that we found by numerical experimentation in the Focus section of Differentiation. The use of Solver can increase accuracy and save a tremendous amount of time by eliminating the need for graphical estimation and numerical experimentation.Differentiation and Solver do have limitations, and cannot be used to replace thought and understanding. For example, the change in marginal cost at a production level of 1,200 thousand drives gives P(q) a local minimum at exactly 1,200. However, the graph of P(q) is not smooth at 1,200. Since it has an angled corner, P(1,200) does not exist. If Solver is constrained to look near 1,200, it cannot find a value of q such that P(q) =0.(materialcontinues)ITClass ProjectMarketing Focus.xlsmC132Using Solver, Focus


MarketingComputerDrivesDue to the lack of sensitivity of profit to production level, Solver can only approximate the level that produces the local minimum 1,200. If it is constrained to look only between 1,190 and 1,210, given an initial value of 1,201, and asked to find a value of q that minimizes P(q); Solver returns 1,199.993.Question 6 in the description of the Class Project asks us to find the profit that Card Tech could expect, if they price the drives at $299.99. We use Solver in the computation table at the bottom of the sheet Marginal. This shows that D(q) = $299.99, when q = 1,176.693, and that the profit from selling 1,176,693 drives will be $41,779,000.Questions 7 and 8 from the Class Project ask how much Card Tech should pay for an advertising campaign that would increase demand for the 12-GB drives by 10% at all price levels. We go to the sheet Functions in Marketing Focus.xlsm. Setting the Demand Factor in Cell F11 at 1.10 increases the sales in all of the test markets by 10%. When the sheet is recalculated, new coefficients for the trend line are displayed in the plot of the demand data. These must by copied into Cells H28:H30.IT(materialcontinues)Class ProjectMarketing Focus.xlsmC133Using Solver, Focus

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