Web Chap A

Web Chapter A Optimization Techniques Overview • Unconstrained & Constrained Optimization

• Calculus of one variable • Partial Differentiation in Economics • Appendix to Web Chapter A: » Lagrangians and Constrained Optimization 2002 South-Western Publishing

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Optimum Can Be Highest or Lowest      • Finding the maximum flying range for the Stealth Bomber is an optimization                problem.   • Calculus teaches that when the first                     derivative is zero, the solution is at an optimum.   • The original Stealth Bomber study showed that a  controversial flying V­wing design optimized the  bomber's range, but the original researchers failed to   find that their solution in fact minimized the range.   • It is critical that managers make decision that     maximize,  not minimize, profit potential! Slide 2

Unconstrained Optimization  • Unconstrained Optimization is a relatively  simple calculus problem that can be  solved using differentiation, such as  finding the quantity that maximizes profit  in the function: 

π (Q) = 16∙Q ­ Q2.  • The answer is Q = 8 as we will see. Where  dπ/dQ = 0.  Slide 3

Constrained Optimization

• Constrained Optimization involves one or more  constraints of money, time, capacity, or energy.   • When there are inequality constraints (as when  you must spend less than or equal to your total  income), linear programming can be used.   • Most often, managers know that some  constraints are binding, which means that they  are equality constraints.   » Lagrangian multipliers are used to solve these problems  (which appears in the Appendix to Web Chapter A). Slide 4

Optimization Format • Economic problems require tradeoffs forced on  us by the limits of our money, time, and energy. • Optimization involves an objective function  and one or more constraints , b. 

Maximize    y = f(x1 , x2 , ..., xn ) Subject to g(x1 , x2 , ..., xn ) < b  or:    Minimize      y = f(x1 , x2 , ..., xn )   Subject to g(x1 , x2 , ..., xn ) > b  Slide 5

Using Equations • profit = f(quantity) or Π = f(Q) » dependent variable & independent variable(s) » average profit = Π/Q » marginal profit = ∆Π / ∆Q

• Calculus uses derivatives » dΠ/dQ = lim ∆Π / ∆Q ∆Q

0

» SLOPE = MARGINAL = DERIVATIVE » NEW DECISION RULE: To maximize profits, find where dΠ/dQ = 0 -- first order condition Slide 6

Quick Differentiation Review Name

Function

Derivative

Example

• Constant Y = c

dY/dX = 0

Y=5 dY/dX = 0

• Line Y = c•X

dY/dX = c

Y = 5•X dY/dX = 5

• Power Y = cXb

dY/dX = b•c•X b-1

Y = 5•X2 dY/dX = 10•X Slide 7

Quick Differentiation Review • Sum Rule Y = G(X) + H(X) example

Y = 5•X + 5•X2

dY/dX = dG/dX + dH/dX dY/dX = 5 + 10•X

• Product Rule Y = G(X)•H(X)

dY/dX = (dG/dX)H + (dH/dX)G

example

Y = (5•X)(5•X2 ) dY/dX = 5(5•X2 ) + (10•X)(5•X) = 75•X2

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Quick Differentiation Review • Quotient Rule Y = G(X) / H(X) dY/dX = (dG/dX)•H - (dH/dX)•G H2 Y = (5•X) / (5•X2) dY/dX = 5(5•X2) -(10•X)(5•X) (5•X2)2 = -25X2 / 25•X4 = - X-2 • Chain Rule Y = G [ H(X) ] dY/dX = (dG/dH)•(dH/dX) Y = (5 + 5•X)2 dY/dX = 2(5 + 5•X)1(5) = 50 + 50•X Slide 9

Applications of Calculus in Managerial Economics • max imizati on

probl em :   

A  profit  function  might  look  like  an  arch,  rising  to  a  peak  and  then  declining  at  even  larger  outputs.    A  firm  might  sell  huge  amounts at very low prices, but discover that profits are low  or negative. • At the maximum, the slope of the profit function is zero.  The  first order condition for a maximum is that the derivative at  that point is zero.   • If  π =  50∙Q  ­  Q2,  then  dπ/dQ  =  50  ­  2∙Q,  using  the  rules  of  differentiation.   • Hence, Q = 25 will maximize profits where 50 ­ 2•Q = 0. Slide 10

More Applications of Calculus • minimization problem: Cost minimization 

supposes that there is a least cost point to produce.  An  average cost curve might have a U­shape.  At the least  cost point, the slope of the cost function is zero.  

• The first order condition for a minimum is that  the derivative at that point is zero.   • If C = 5∙Q2 ­ 60∙Q, then dC/dQ = 10∙Q ­ 60.   • Hence, Q = 6 will minimize cost where    10•Q ­ 60 = 0. Slide 11

More Examples • Competitive Firm: Maximize Profits » where Π = TR - TC = P•Q - TC(Q) » Use our first order condition: dΠ /dQ = P - dTC/dQ = 0. a function of Q » Decision Rule: P = MC. Problem 1

●

Max Π = 100•Q - Q2 100 -2•Q = 0 implies Q = 50 and Π = 2,500

Problem 2

●

Max Π = 50 + 5•X2 So, 10•X = 0 implies Q = 0 and Π = 50 Slide 12

Second Order Condition: One Variable

• If the second derivative is negative, then it’s a maximum • If the second derivative is positive, then it’s a minimum Problem 1 ●Max

Problem 2

Π = 100•Q - Q

2

100 -2•Q = 0 second derivative is: -2 implies Q =50 is a MAX

●Max

Π = 50 + 5•X2

10•X = 0 second derivative is: 10 implies Q = 0 is a MIN Slide 13

Partial Differentiation • Economic relationships usually involve several independent variables. • A partial derivative is like a controlled experiment -- it holds the “other” variables constant • Suppose price is increased, holding the disposable income of the economy constant as in Q = f (P, I ), then ∂Q/∂P holds income constant. Slide 14

Problem: • Sales are a function of advertising in newspapers and magazines ( X, Y) • Max S = 200X + 100Y -10X2 -20Y2 +20XY • Differentiate with respect to X and Y and set equal to zero.

∂S/∂X = 200 - 20X + 20Y= 0 ∂S/∂Y = 100 - 40Y + 20X = 0 • solve for X & Y and Sales Slide 15

Solution: 2 equations & 2 unknowns • 200 - 20X + 20Y= 0 • 100 - 40Y + 20X = 0 • Adding them, the -20X and +20X cancel, so we get 300 - 20Y = 0, or Y =15 • Plug into one of them: 200 - 20X + 300 = 0, hence X = 25 • To find Sales, plug into equation: S = 200X + 100Y -10X2 -20Y2 +20XY = 3,250 Slide 16

International Import Restraints • Import quotas of Japanese automobiles are  inequality constraints. The added constraint will  affect decisions. • A Japanese manufacturer will shift more  production to U.S. assembly facilities and  increase the price of cars exported to the U.S.   • We may also expect that the exported cars will be  "top of the line" models, and we expect U.S.  manufacturers to raise domestic car prices. Slide 17

Web Chapter A -- Appendix Objective functions are often constrained by one or more “constraints” (time, capacity, or money) Max L = (objective fct.) - λ{constraint set to zero} Min L = (objective fct.) +λ{constraint set to zero} An artificial variable is created for each  constraint in the Lagrangian multiplier technique.   This artificial variable is traditionally called  Slide 18

Maximize Utility Example example: Max Utility subject to a money constraint Max U = X•Y2 subject to a $12 total budget with the prices of X as $1, the price of Y as $4 (suppose X represents soda and Y movie tickets). Max L = X•Y2 - λ { X + 4Y - 12} • differentiate with respect to X, Y and lambda, λ. Slide 19

∂L/∂X = Y2 - λ = 0 ∂L/∂Y = 2XY - 4λ = 0 ∂L/∂λ = X + 4Y- 12 = 0

Y2 = λ 2XY = 4λ

Three equations and three unknowns Solve: Ratio of first two equations is: Y/2X = 1/4 or Y = .5 X. Substitute into the third equation: We get:

X = 4; Y = 2; and λ = 4 • Lambda is the marginal (objective function) of the (constraint). In the parentheses, substitute the words used for the objective function and constraint.

• Here, λ = the marginal utility of money.

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Problem Minimize crime in your town • • • •

Police, P, costs $15,000 each. Jail, J, costs $10,000 each. Budget is $900,000. Crime function is estimated: C = 5600 - 4PJ » Set up the problem as a Lagrangian » Solve for optimal P and J, and C » What is economic meaning of lambda? Slide 21

Answer • Min L= 5600 - 4PJ + λ{15,000•P + 10,000•J -900,000 } • To Solve, differentiate

1. ∂L/∂P: - 4•J +15,000•λ = 0 2. ∂L/∂J: - 4•P +10,000•λ = 0 3. ∂L/∂λ : 15,000•P +10,000•J -900,000 =0 J/P = 1.5 so J = 1.5•P & substitute into (3.) 15,000•P +10,000•[1.5•P] - 900,000 = 0 solution: P = 30, J = 45, C = 200 and λ = -.012 • Lambda is the marginal crime (reduction) for a dollar of additional budget spent

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