Final Exam for 03-62-194: Mathematics for Business

3:30 - 6:30 P.M., December 16, 2011

Solutions

1. (4 marks) A company charting its profits notices that the relationship between thenumber of units sold x and the profit y is linear. If 200 units are sold the profit is$3, 100. If 250 units are sold the profit is $6, 000. Give an equation for the profit y as afunction of the number of units x.

Solution. The slope is m = (6000 − 3100)/(250 − 200) = 58. When x = 200, y = 3100so that 3100 = 58(200) + b giving b = −8500. An equation is y = 58x− 8500.

2. (5 marks) A manufacturer of a DVD player has a fixed cost of $9, 800 and variable costsof $65 per unit. The company sells each DVD for $100. Let x be the number of unitsproduced and sold.

(a) (1 mark) Give an equation for the cost function C(x).

(b) (1 mark) Give an equation for the revenue function R(x).

(c) (1 mark) Give an equation for the profit function P (x).

(d) (2 marks) How many units must be sold to break even?

Solution. We have

(a) C(x) = 65x + 9800.

(b) R(x) = 100x.

(c) P (x) = 35x − 9800.

(d) We solve P (x) = 0 to get x = 280.

3. (5 marks) The demand for a certain product is given by 2p + 5q = 200 and the supplyis given by p − 2q = 10, where p is the market price in dollars, and q is the quantitydemanded or supplied.

(a) (4 marks) Find the equilibrium price and equilibrium quantity.

(b) (1 mark) What quantity is supplied and demanded if the price is $60?

Solution. We solve the two equations to get p = 50 and q = 20. When p = 60 thedemand is q = 16 and the supply is q = 25.

4. (7 marks) Solve the following system by creating the associated augmented matrix andthen reducing it to echelon form.

x− 2y + 3z = 4

2x + 3y − z = 1

x + 2y − 3z = −6

Solution. We have the following sequence of row equivalent matrices

[A|b] =

1 −2 3 42 3 −1 11 2 −3 −6

R2←→ R2− 2R1R3←→ R3− R1

1 −2 3 40 7 −7 −70 4 −6 −10

R2←→ R2/7

1 −2 3 40 1 −1 −10 4 −6 −10

R1←→ R1 + 2R2

R3←→ R3− 4R2

1 0 1 20 1 −1 −10 0 −2 −6

R3←→−R3/2

1 0 0 −10 1 0 20 0 1 3

The answer is x = −1, y = 2, and z = 3.

5. (4 marks) An investor has $235, 000 invested in three income properties. The first prop-erty earns 12%, the second earns 10% and the third earns 8% per year. The investor’sannual income from the three properties is $22, 500. The amount invested at 8% is twicethat invested at 12%.

Set up, but do not solve, the system of linear equations whose solution will

give the amount invested in each property. One mark will be given for a

well-written definition of the variables.

Solution. Let x, y and z be the amount of money, in dollars, invested in the first,second, and third properties, etc. The equations are

x + y + z = 235, 000

.12x + .10y + .08z = 22, 500

2x− z = 0

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6. (6 marks) Two linear programming problems have the same bounded, non-empty feasibleregion. The corner points of the region are (0, 0), (7, 0), (6, 5), (4, 5), and (0, 3).

(a) (3 marks) What is the maximum value of C = −x− 2y?

(b) (3 marks) What is the minimum value of Z = 5x− 3y?

Solution.Corner Point C = −x− 2y Z = 5x− 3y

(0, 0) 0 0(7, 0) −7 35(6, 5) −16 15(4, 5) −14 5(0, 3) −6 −9

The maximum value of C = −x− 2y is 0. The minimum value of Z = 5x− 3y is −9.

7. (6 marks) Solve the following system graphically and clearly indicate the solution.

−2x + 5y ≤ 206x + 5y ≤ 60x ≥ 0

y ≥ 0

Solution. The corner points (0, 0), (0, 4), (10, 0) and (5, 6).

8. (7 marks) Your job is to make, at minimum cost, a mixture of nuts from almonds,cashews, and peanuts for an upcoming party. The almonds cost $2.50/kg, the cashewscost $4.00/kg and the peanuts cost $1.50/kg. There are some restrictions. The mixtureof nuts must weigh 10kg or more. The combined weight of almonds and cashews mustbe greater than or equal to the weight of the peanuts. The weight of the cashews mustbe 1kg or more. The weight of almonds must be less than or equal to the weight of thecashews.

Set up, but do not solve the LP. Be sure to include a well written definition

of the variables, which is worth 1 of the seven marks. Solution.

Let A, C , and P be the number of kilograms of almonds, cashews, and peanuts to bepurchased. The linear programming problem is to

min 2.5A + 4C + 1.5P = zsubject to A + C + P ≥ 10

A + C − P ≥ 0C ≥ 1

A − C ≤ 0A ≥ 0

C ≥ 0P ≥ 0

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9. (5 marks) Graph the quadratic function y = x2 − 6x + 5. Give the coordinates of thevertex, the y-intercept and the two x-intercepts. Clearly label these points on the graph.

Solution. The vertex is (3,−4), the y-intercept is (0, 5) and the x-intercepts are (1, 0)and (5, 0).

10. (5 marks) What is the future value of $3, 500 invested at 3% compounded semi-annuallyafter 4 years?

Solution. The future value is FV = 3500(1 + .032

)4(2) = 3500(1.015)8 = 3942.72.

11. (5 marks) What is the future value of $1, 000 invested at 2% compounded continuouslyafter 5 years?

Solution. The future value is FV = 1000e5(.02) = 1000e.1 = 1105.17

12. (5 marks) If interest is 4% compounded monthly, what is the present value of $500 threeyears from now?

Solution. The present value is PV = 500(1+ .04

12)3(12)

= 443.55.

13. (5 marks) If interest is 3.5% compounded continuously, what is the present value of$10, 000 six years from now?

Solution. The present value is PV = 10000e.035(6) = 8105.84.

14. (6 marks) An advertising agency has found that when it promotes a new product in amarket with a population of 210 000 the number of people who are aware of this productt weeks after the ad campaign starts is given by

f(t) = 210, 000(1 − e−.045t).

(a) (2 marks) How many people are aware of the product after 2 weeks?

(b) (4 marks) How many weeks will it take until half the people in this market areaware of the new product?

Solution. (a) We have f(2) = 210, 000(1− e−.045∗2) = 18074.45. The number is 18, 074.

(b) We solve 105, 000 = 210, 000(1 − e−.045∗t) to get t = ln(.5)−.045

= 15.4. It takes just over15 weeks.

15. (4 marks) An insurance company received 25 claims. It is known that 2 of the claimsare fraudulent. If two claims are chosen at random for investigation, find the probabilitythat

(a) (2 marks) both fraudulent claims are investigated.

(b) ( 2 marks) at most one fraudulent claim is investigated.

Solution. There is one way to pick both fraudulent claims and there are 25C2 = 300ways of picking two cases from twenty-five. The probability is 1/300. For part (b) wehave probability 299/300.

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16. (4 marks) A product goes through two different kinds of inspection; one for appearanceand one for functionality. The probability that the product passes the appearance in-spection is 0.45. The probability that it passes the appearance inspection and not thefunctionality inspection is 0.3. What is the probability that it will pass the appearanceinspection and the functionality inspection?

Solution. P (A) = P (A ∩ B) + P (A ∩ not B). We have .45 = P (A ∩ B) + .3 so thatP (A ∩B) = .15.

17. (5 marks) NACO Body Shops has found that 14% of the cars on the road need to bepainted and that 10% need major body work. If a car needs major body work, there isa 20% chance that it also needs to be painted.

(a) (2 marks) What is the probability that a randomly chosen car needs to be paintedand it needs major body work?

(b) (3 marks) What is the probability that a randomly chosen car needs major bodywork, given that it needs to be painted?

Solution. Let A be the event of needing paint and let B be the event of needing bodywork. We have P (A) = .14, P (A|B) = .2 and P (B) = .1.

(a) Pr(A ∩B) = Pr(A|B)Pr(B) = (.2)(.1) = .02.

(b) Pr(B|A) = Pr(A ∩ B)/Pr(A) = .02/.14 = .143.

18. (4 marks) An experiment has four outcomes with value and probabilities given in thetable.

i 1 2 3 4

Value xi 4 2 8 3Probability P (xi) .2 .1 .3 .4

(a) (2 marks) What is the expected value µ?

(b) (3 marks) What is the standard deviation σ?

Solution. The expected value is

µ = x1P (x1) + x2P (x2) + x3P (x3) + x4P (x4)= 4.6

The variance is

σ2 = (x1 − µ)2P (x1) + (x2 − µ)2P (x2) + (x3 − µ)2P (x3) + (x4 − µ)2P (x4)= 5.24.

The standard deviation is√

5.24 ' 2.29.

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19. (3 marks) A hospital buys thermometers in boxes of 1000. The probability that athermometer is defective is 0.01. If 20 thermometers are selected from one box, what isthe probability that exactly 2 are defective?

Solution. This is a binomial experiment with p = .01, q = .99, n = 20 and x = 2. Theprobability is

P (2) = 20C2(.01)2(.99)18 =20!

18!2!(.01)2(.99)18 ' .0159.

Solution.

20. (5 marks) In a certain store, there is a 0.02 probability that the scanned price in the barcode scanner will not match the advertised price. The cashier scans 2000 items. Whatis the probability of 30 or more mismatches?

Solution. We use the normal approximation to the binomial distributions. The meanis µ = 2000(.02) = 40 and the standard deviation is σ =

√

2 000(.02)(.98) ' 6.26. Theprobability of 30is approximated by

P (X ≥ 29.5) = 1−NORMDIST (29.5, 40, 6.26, 1).

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