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# Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] - PowerPoint PPT Presentation

Chabot Mathematics. §3.2b System Applications. Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] MTH 55. 3.2. Review §. Any QUESTIONS About §3.2a → System Applications Any QUESTIONS About HomeWork §3.2a → HW-09. Summary of Eqn Elimination.

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§3.2b SystemApplications

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]

3.2

Review §

• §3.2a → System Applications

• §3.2a → HW-09

• Arrange the equations with like terms in columns.

• Multiply one or both equations by an appropriate factor so that the new coefficients of x or y have the same absolute value.

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

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

• Check the solution in each of the original equations.

Example  Solve by Elimination

• → from the top equation eliminating the variable x

• 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

Example  Eqn Elimination cont.1

• Choose to Find y first; using Algebra:

Subtract the bottom equation from the top equation.

Divide Both Sides by 14

Solve for y

Example  Eqn Elimination cont.2

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

Combine Like Terms

Mult Both Sides by 1/3

Solve for x

Example  Eqn Elimination cont.3

• Thus the Solution to the Eqn System

• To Check, Substitute the value of the variables into each original equation

Example  Graphically

(5,3)

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

Translating

• The Two Keys to the Translation

• Use the LET Statement to ASSIGN VARIABLES (Letters) to Unknown Quantities

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

Example  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  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

Example  Boat in Motion

• 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.

Example  Boat in Motion

• 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.

Example  Boat in Motion

• 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

Example  Boat in Motion

• The Table produced an Eqn System

Solve: Use Substitution

(r + 5)4= (r – 5)6

4r + 20= 6r – 30

50= 2r

25= r

Example  Boat in Motion

• 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.

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

Example  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?

Example  Autos in Motion

• 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

• Translate: LET

• x≡ Tamika’s travel time after passing Exit 54

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

Example  Autos in Motion

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

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

Example  Autos in Motion

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

• From Table

Carry out: The translations produced a System of Eqns

Example  Autos in Motion

• Solve by Substitution

• Sub y = 0.92 into firstEqn to solve for x

Example  Autos in Motion

• Check: Both Eqns

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

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

Equilibrium point

Quantity

Demand

Price

Supply & Demand Equilibrium

• 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  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

supply

demand

• Where

• p is the unit-price in dollars

• x is the number of units

Example  Supply = Demand

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

• Translate: S = D →

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

Example  Supply = Demand

• Subbing

Example  Supply = Demand

• To find the price pback-substitute x = 10,000 into the Supply Eqn

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

• The Graph serves as the Chk

• 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.

• Total profit is the money taken in less the money spent, or total revenueminus 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)

• There are two types of costs.

• Costs which must be paid whether a product is produced or not, are called fixed costs.

• 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  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.

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

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

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

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

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

Example  Profit

• SOLUTION

• 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.

• Total revenue is given by

• R(x) = 10x

• \$10 times the no. of wallets sold

Example  Profit

• SOLUTION

• Total profit is given by

• P(x) = R(x) – C(x)

• = 10x – (2,400 + 2x)

• = 8x – 2,400.

• Total profit for 500 units

• P(500) = 8(500) – 2,400

• = 4,000 – 2,400

• = \$1,600.

R(x) = 10x

Break-even point

Gain

4,000

3,500

3,000

C(x) = 2400 + 2x

2,500

2,000

Loss

1,500

P(x) = 8x – 2400

1,000

500

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

Wallets sold

-2500

Example  Profit

• The graphs of R(x), C(x), and P(x) in \$

Example  Break-Even Point

• In the Previous Examplenote the Cost and Revenue functions

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

• Find BreakEven for the Wallet Factory

Example  Break-Even Point

• CarryOut

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

• Problems From §3.2 Exercise Set

• 38, 50

• WashableWallet

Supply & DemandSeeSaw

Appendix

Bruce Mayer, PE

Licensed Electrical & Mechanical [email protected]