1. Slide 1  2005 South-Western Publishing Appendix 2A Differential Calculus in Management A function with one decision variable, X, can be written as Y = f(X) The marginal value of Y, with a small increase of X, is M y =  Y/  X For a very small change in X, the derivative is written: dY/dX = limit  Y/  X  X  B

2. Slide 2 Marginal = Slope = Derivative The slope of line C-D is  Y/  X The marginal at point C is M y is  Y/  X The slope at point C is the rise (  Y) over the run (  X) The derivative at point C is also this slope X C D Y YY XX

3. Slide 3 Optimum Can Be Highest or Lowest Finding the maximum flying range forthe 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!

4. Slide 4 Quick Differentiation Review Constant Y = cdY/dX = 0Y = 5 Functions dY/dX = 0 A Line Y = cXdY/dX = cY = 5X dY/dX = 5 Power Y = cX b dY/dX = bcX b-1 Y = 5X 2 Functions dY/dX = 10X Name Function Derivative Example

5. Slide 5 Sum of Y = G(X) + H(X) dY/dX = dG/dX + dH/dX Functions exampleY = 5X + 5X 2 dY/dX = 5 + 10X Product of Y = G(X)H(X) Two Functions 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

6. Slide 6 Quotient of Two Y = G(X) / H(X) Functions 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

7. Slide 7 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.

8. Slide 8 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.

9. Slide 9 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

10. Slide 10 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

11. Slide 11 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.

12. Slide 12 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

13. Slide 13 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