1. Slide 1  2002 South-Western Publishing 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

2. Slide 2 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 derivativeis 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!

3. Slide 3 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  /dQ = 0.

4. Slide 4 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).

5. Slide 5 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,..., x n ) S ubject to g(x 1, x 2,..., x n ) < b or: Minimize y = f(x 1, x 2,..., x n ) Subject to g(x 1, x 2,..., x n ) > b

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

7. Slide 7 Quick Differentiation Review Constant Y = cdY/dX = 0Y = 5 dY/dX = 0 Line Y = cXdY/dX = cY = 5X dY/dX = 5 Power Y = cX b dY/dX = bcX b-1 Y = 5X 2 dY/dX = 10X Name Function Derivative Example

8. Slide 8 Sum Rule Y = G(X) + H(X) dY/dX = dG/dX + dH/dX exampleY = 5X + 5X 2 dY/dX = 5 + 10X Product Rule Y = G(X)H(X) dY/dX = (dG/dX)H + (dH/dX)G example Y = (5X)(5X 2 ) dY/dX = 5(5X 2 ) + (10X)(5X) = 75X 2 Quick Differentiation Review

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

10. Slide 10 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  /dQ = 50 - 2·Q, using the rules of differentiation. Hence, Q = 25 will maximize profits where 50 - 2Q = 0.

11. Slide 11 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 dC/dQ = 10·Q - 60. Hence, Q = 6 will minimize cost where 10Q - 60 = 0.

12. Slide 12 More Examples Competitive Firm: Maximize Profits »where  = TR - TC = PQ - TC(Q) »Use our first order condition: d  /dQ = P - dTC/dQ = 0. »Decision Rule: P = MC. a function of Q Max  = 100Q - Q 2 100 -2Q = 0 implies Q = 50 and  = 2,500 Max  = 50 + 5X 2 So, 10X = 0 implies Q = 0 and  = 50 Problem 1Problem 2

13. Slide 13 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 Max  = 100Q - Q 2 100 -2Q = 0 second derivative is: -2 implies Q =50 is a MAX Max  = 50 + 5X 2 10X = 0 second derivative is: 10 implies Q = 0 is a MIN Problem 1Problem 2

14. Slide 14 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.

15. Slide 15 Problem: Sales are a function of advertising in newspapers and magazines ( X, Y) Max S = 200X + 100Y -10X 2 -20Y 2 +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

16. Slide 16 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 -10X 2 -20Y 2 +20XY = 3,250

17. Slide 17 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.

18. Slide 18 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,.

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

20. Slide 20  L/  X = Y 2 -  = 0 Y 2 =   L/  Y = 2XY - 4  = 0 2XY = 4   L/  = X + 4Y- 12 = 0 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  = 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.

21. Slide 21 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?

22. Slide 22 Answer Min L= 5600 - 4PJ + {15,000P + 10,000J -900,000 } To Solve, differentiate  L/  P: - 4J +15,000   L/  J: - 4P +10,000   L/  : 15,000P +10,000J -900,000 =0 J/P = 1.5 so J = 1.5P & substitute into (3.) 15,000P +10,000[1.5P] - 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