bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

[email protected] • MTH55_Lec-12_sec_3-2b_Eqn_Sys_Apps.ppt1

Bruce Mayer, PE Chabot College Mathematics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

§3.2b System§3.2b SystemApplicationsApplications



Review §Review §

Any QUESTIONS About• §3.2a → System Applications

Any QUESTIONS About HomeWork• §3.2a → HW-09

3.2 MTH 55



Summary of Eqn EliminationSummary of Eqn Elimination1. Arrange the equations with

like terms in columns.2. Multiply one or both equations by an

appropriate factor so that the new coefficients of x or y have the same absolute value.

3. Add or subtract the equations and solve for the remaining variable.

4. Substitute the value for that variable into one of the equations and solve for the value of the other variable.

5. Check the solution in each of the original equations.



Example Example Solve by Solve by EliminationElimination Consider This

Equation Systemx y

x y

4 17

3 2 9 If the top equation

was multiplied by 3, then the first term would be 3x. The bottom equation could then be subtracted →

Then the New System of Eqns

3 4 3 17

3 12 51

( ) ( )x y

x y

3 12 51

3 2 9

x y

x y

→ from the top equation eliminating the variable x



Example Example Eqn Elimination Eqn Elimination cont.1 cont.1

Choose to Find y first; using Algebra:

3 12 51

3 2 9

x y

x y

14 42y

14

14

42

14

y

y 3

Subtract the bottom equation from the top equation.

Divide Both Sides by 14

Solve for y




Solve for x by substituting the value for y (y = 3) into one of the equations.

Add 6 to Both Sides

Combine Like Terms

3 2 9x y 3 2 3 9x ( )

69663 x

153 x

31153 x Mult Both Sides by 1/3

5x Solve for x




Thus the Solution to the Eqn System

5x y 3

To Check, Substitute the value of the variables into each original equationx y 4 17

5 4 3 17 ( )

5 12 17 17 17

3 2 9x y 3 5 2 3 9( ) ( )

15 6 9 9 9

x y

x y

4 17

3 2 9



Example Example Graphically Graphically

-3

-2

-1

0

1

2

3

4

5

6

7

-1 0 1 2 3 4 5 6 7 8 9

file = M65_§7-1_Graphs_0607.xls

(5,3)

923 yx

174 yx



The Problem Solving CycleThe Problem Solving Cycle



Applications TipsApplications Tips

The Most Important Part of Solving REAL WORLD (Applied Math) Problems

The Two Keys to the Translation• Use the LETLET Statement to ASSIGN

VARIABLES (Letters) to Unknown Quantities

• Analyze the RELATIONSHIP Among the Variables and Constraints (Constants)



Example Example Motion Problems Motion Problems Recall the Motion Formula:

Distance = {Rate (or speed)} {Time}• Symbolically → d = r•t

Game plan for Solving Motion Probs• Draw a diagram using an arrow or arrows to

represent distance and the direction of each object in motion.

• Organize the information in a table or chart.• Look for as many things as you can that are

the same, so you can write equations



Example Example Boat in Motion Boat in Motion

Miguel’s motorboat took 4 hr to make a trip downstream with a 5-mph current. The return trip against the same current took 6 hr. Find the speed of the boat in still water.

AQUICKDiagram




1. Familiarize. Note that the current speeds up the boat when going downstream, but slows down the boat when going upstream. For our guess, suppose that the speed of the boat with no current is 20 mph.The boat would then travel

• 20 + 5 = 25 mph downstream

• 20 – 5 = 15 mph upstream.




Assess 20 mph Guess• In 4 hr downstream the boat would travel

4(25) = 100 mi.

• In 6 hr upstream the boat would travel 6(15) = 90 mi.

So our guess of 20 mph is incorrect, but it seems pretty close.




2. Translate: LET r ≡ the rate of the boat in still water.

Then r + 5 = the boat’s speed downstream r – 5 = the boat’s speed upstream.

Tabulate d = r•t calculations

6r – 5dUpstream

4r + 5dDownstream

TimeRateDistance

d = (r + 5)4

d = (r – 5)66r – 5dUpstream

4r + 5dDownstream

TimeRateDistance

6r – 5dUpstream

4r + 5dDownstream

TimeRateDistance

d = (r + 5)4

d = (r – 5)6




The Table produced an Eqn System

( 5)4,

( 5)6.

d r

d r

(r + 5)4 = (r – 5)6

4r + 20 = 6r – 30

50 = 2r

25 = r

3. Solve: Use Substitution




4. Check: When r = 25 mph, the speed downstream is 30 mph and the speed upstream is 20 mph. The distance downstream is 30(4) = 120 mi and the distance upstream is 20(6) = 120 mi, so we have a check.

5. State: The speed of the boat in still water is 25 mph.



Example Example Autos in Motion Autos in Motion

Tamika and Ernesto are traveling north in separate cars on the same highway. Tamika is traveling at 55 miles per hour and Ernesto is traveling at 70 miles per hour. Tamika passes Exit 54 at 1:30 p.m. Ernesto passes the same exit at 1:45 p.m.

At what time will Ernesto catch up with Tamika?




1. Familiarize: To determine what time Ernesto will catch up with Tamika, we need to calculate the amount of time it will take him to catch up to her. We can then add the amount to 1:45pm

2. Translate: LET• x ≡ Tamika’s travel time after

passing Exit 54

• y ≡ Ernesto’s travel time after passing Exit 54




Tabulate (distance) = (spd)(time) calcs

Category Speed Time Distance

Tamika 55 x 55x

Ernesto 70 y 70y

Translate – Connect the Time Difference: Ernesto reaches exit-54 15 minutes after Tamika; Tamika will have traveled 15 minutes (¼ hr) longer → 4

1yx




Translate: When Ernesto catches up, they will have traveled the same distance; i.e.; dTamika = dErnesto.

• From Table

3. Carry out: The translations produced a System of Eqns

yx 7055 1

455 70

x y

x y




Solve by Substitution 1

455 70

x y

x y

155 70

4

y y

5555 70

4 y y

13.75 15 y

0.92 y

Sub y = 0.92 into firstEqn to solve for x

1

40.92 0.25

x y

x

1.17x




4. Check: Both Eqns

55 70

55(1.17) 70(0.92)

64.35 64.4

x y

1

41.17 0.92 0.25

1.17 1.17

x y

5. State: Ernesto will catch up to Tamika in a little less than 1 hour (0.92 hrs, which is 55 min). The time will be 1:45pm + (55 min) = 2:40pm



Supply and DemandSupply and Demand

As the price of a product varies, the amount sold varies. Consumers will demand less as price goes up. Sellers will supply more as the price goes up.



Supply & Demand EquilibriumSupply & Demand EquilibriumSupply

Demand

Equilibrium point

Price

Qua

ntit

y

The point of intersection is called the equilibrium point. At that price, the amount that the sellers will supply is the same amount that the consumers will buy



Example Example Supply = Demand Supply = Demand

Find the equilibrium point if the supply and demand functions for a new brand of digital video recorder (DVR) are given by the system

p 60 0.0012x (1)

p 80 0.0008x (2) Where

• p is the unit-price in dollars

• x is the number of units

supply

demand




1. Familiarize: The word “Equilibrium” in Supply & Demand problems means that Supply & Demand are Exactly the Same.

2. Translate: S = D →

SD pxxp 0120600008080 ..

3. Carry Out: Isolate x in the above Eqn to find the quantity. Then Substitute the value of x into either Price Eqn




Subbing p 80 0.0008x

60 0.0012x 80 0.0008x

0.0012x 20 0.0008x

0.002x 20

x 20

0.002x 10,000




To find the price p back-substitute x = 10,000 into the Supply Eqn

p 60 0.0012x

60 0.0012 10,000 72

5. State: The equilibrium point is (10,000, 72).

• The Graph serves as the Chk



Break Even AnalysisBreak Even Analysis When a company manufactures x units

of a product, it spends money. This is total cost and can be thought of as a function C, where C(x) is the total cost of producing x units. When a company sells x units of the product, it takes in money. This is total revenue and can be thought of as a function R, where R(x) is the total revenue from the sale of x units.



Profit FunctionProfit Function

Total profit is the money taken in less the money spent, or total revenue minus total cost.

Total profit from the production and sale of x units is a function P given by

Profit = Revenue – Cost

or

P(x) = R(x) – C(x)



Production CostProduction Cost

There are two types of costs. 1. Costs which must be paid whether a

product is produced or not, are called fixed costs.

2. Costs that vary according to the amount being produced are called variable costs.

The sum of the fixed cost and variable cost gives the total cost.



Example Example Profit Profit A specialty wallet company has fixed costs that

are $2,400. Each wallet will cost $2 to produce (variable costs) and will sell for $10.

a) Find the total cost C(x) of producing x wallets.

b) Find the total revenue R(x) from the sale of x wallets.

c) Find the total profit P(x) from the production and sale of x wallets.

d) What profit will the company realize from the production and sale of 500 wallets?

e) Graph the total-cost, total-revenue, and total-profit functions. Determine the break-even point.



Example Example Profit Profit

SOLUTIONa) Total cost is given by

• C(x) = (Fixed costs) plus (Variable costs)

• C(x) = 2,400 + 2x.– where x is the number of wallets produced.

b) Total revenue is given by

• R(x) = 10x – $10 times the no. of wallets sold



Example Example Profit Profit

SOLUTIONc) Total profit is given by

• P(x) = R(x) – C(x)

• = 10x – (2,400 + 2x)

• = 8x – 2,400.

d) Total profit for 500 units

• P(500) = 8(500) – 2,400

• = 4,000 – 2,400

• = $1,600.



Example Example Profit Profit The graphs

of R(x), C(x), and P(x) in $

0 50 100 150 200 250 300 350 400 450 500 550

500

1,5001,000

2,5002,000

3,000

4,0003,500

R(x) = 10x

C(x) = 2400 + 2x

P(x) = 8x – 2400

-2500

Break-even point

Loss

Gain

Wallets sold



Example Example Break-Even Point Break-Even Point

In the Previous Examplenote the Cost and Revenue functions

xxR

xxC

10

24002

At BreakEven the Revenues just barely cover the Costs; i.e., R(xBE) = C(xBE)

Find BreakEven for the Wallet Factory

xRxxxC BEBEBE 1024002



Example Example Break-Even Point Break-Even Point

CarryOut

30082400

24008

2400210

1024002

BE

BE

BEBE

BEBE

x

x

xx

xx

Thus this Wallet production operation becomes Profitable at production levels of greater than 300 wallets



WhiteBoard WorkWhiteBoard Work

Problems From §3.2 Exercise Set• 38, 50

WashableWallet



All Done for TodayAll Done for Today

Supply &

DemandSeeSaw



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Chabot Mathematics

AppendiAppendixx

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srsrsr 22

bruce mayer, pe licensed electrical & mechanical engineer bmayer@chabotcollege

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