Web Chapter A Optimization Techniques Overview Unconstrained Constrained

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 Slide 1

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 V-wing 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: (Q) = 16·Q - Q 2. • The answer is Q = 8 as we will see. Where d /d. Q = 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(x 1 , x 2 , . . . , xn ) Subject to g(x 1 , x 2 , . . . , xn ) < b or: Minimize y = f(x 1 , x 2 , . . . , xn ) Subject to g(x 1 , x 2 , . . . , xn ) > b Slide 5

Using Equations • profit = f(quantity) or = f(Q) » dependent variable & independent variable(s) » average profit = Q » marginal profit = / Q • Calculus uses derivatives » d /d. Q = lim / Q Q 0 » SLOPE = MARGINAL = DERIVATIVE » NEW DECISION RULE: To maximize profits, find where d /d. Q = 0 -- first order condition Slide 6

Quick Differentiation Review Name Function Derivative Example • Constant Y = c d. Y/d. X = 0 Y=5 d. Y/d. X = 0 • Line Y = c • X d. Y/d. X = c Y = 5 • X d. Y/d. X = 5 • Power Y = c. Xb d. Y/d. X = b • c • X b-1 Y = 5 • X 2 d. Y/d. X = 10 • X Slide 7

Quick Differentiation Review • Sum Rule Y = G(X) + H(X) example Y = 5 • X + 5 • X 2 d. Y/d. X = d. G/d. X + d. H/d. X d. Y/d. X = 5 + 10 • X • Product Rule Y = G(X) • H(X) d. Y/d. X = (d. G/d. X)H + (d. H/d. X)G example Y = (5 • X)(5 • X 2 ) d. Y/d. X = 5(5 • X 2 ) + (10 • X)(5 • X) = 75 • X 2 Slide 8

Quick Differentiation Review • Quotient Rule Y = G(X) / H(X) d. Y/d. X = (d. G/d. X) • H - (d. H/d. X) • G H 2 Y = (5 • X) / (5 • X 2) d. Y/d. X = 5(5 • X 2) -(10 • X)(5 • X) (5 • X 2)2 = -25 X 2 / 25 • X 4 = - X-2 • Chain Rule Y = G [ H(X) ] d. Y/d. X = (d. G/d. H) • (d. H/d. X) Y = (5 + 5 • X)2 d. Y/d. X = 2(5 + 5 • X)1(5) = 50 + 50 • X Slide 9

Applications of Calculus in Managerial Economics • maximization problem: 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 = 50·Q - Q 2, then d /d. Q = 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 U-shape. 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·Q 2 - 60·Q, then d. C/d. Q = 10·Q - 60. • Hence, Q = 6 will minimize cost where 10 • Q - 60 = 0. Slide 11

More Examples • Competitive Firm: Maximize Profits » where = TR - TC = P • Q - TC(Q) » Use our first order condition: d /d. Q = P - d. TC/d. Q = 0. a function of Q » Decision Rule: P = MC. Problem 1 l Max = 100 • Q - Q 2 100 -2 • Q = 0 implies Q = 50 and = 2, 500 Problem 2 l Max = 50 + 5 • X 2 So, 10 • X = 0 implies Q = 0 and = 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 l. Max = 100 • Q - Problem 2 Q 2 100 -2 • Q = 0 second derivative is: -2 implies Q =50 is a MAX l. Max = 50 + 5 • X 2 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 = 200 X + 100 Y -10 X 2 -20 Y 2 +20 XY • Differentiate with respect to X and Y and set equal to zero. S/ X = 200 - 20 X + 20 Y= 0 S/ Y = 100 - 40 Y + 20 X = 0 • solve for X & Y and Sales Slide 15

Solution: 2 equations & 2 unknowns • 200 - 20 X + 20 Y= 0 • 100 - 40 Y + 20 X = 0 • Adding them, the -20 X and +20 X cancel, so we get 300 - 20 Y = 0, or Y =15 • Plug into one of them: 200 - 20 X + 300 = 0, hence X = 25 • To find Sales, plug into equation: S = 200 X + 100 Y -10 X 2 -20 Y 2 +20 XY = 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. ) - {constraint set to zero} Min L = (objective fct. ) + {constraint set to zero} An artificial variable is created for each constraint in the Lagrangian multiplier technique. This artificial variable is traditionally called lambda, . Slide 18

Maximize Utility Example example: Max Utility subject to a money constraint Max U = X • Y 2 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 • Y 2 - { X + 4 Y - 12} • differentiate with respect to X, Y and lambda, . Slide 19

L/ X = Y 2 - = 0 L/ Y = 2 XY - 4 = 0 L/ = X + 4 Y- 12 = 0 Y 2 = 2 XY = 4 Three equations and three unknowns Solve: Ratio of first two equations is: Y/2 X = 1/4 or Y =. 5 X. Substitute into the third equation: We get: X = 4; Y = 2; and = 4 • Lambda is the marginal (objective function) of the (constraint). In the parentheses, substitute the words used for the objective function and constraint. • Here, the marginal utility of money. Slide 20

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 - 4 PJ » 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 - 4 PJ + {15, 000 • P + 10, 000 • J -900, 000 } • To Solve, differentiate L/ P: - 4 • J +15, 000 • L/ J: - 4 • P +10, 000 • L/ : 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 = -. 012 • Lambda is the marginal crime (reduction) for a dollar of additional budget spent Slide 22