Slide 12005 South-Western Publishing

Appendix 2ADifferential 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 My = Y/X

• For a very small change in X, the derivative is written:

dY/dX = limit Y/XX B

Slide 2

Marginal = Slope = Derivative

• The slope of line C-D is Y/X

• The marginal at point C is My 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

DYY

X

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!

Slide 4

Quick Differentiation Review

• Constant Y = c dY/dX = 0 Y = 5

Functions dY/dX = 0

• A 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

Functions dY/dX = 10•X

Name Function Derivative Example

Slide 5

• Sum of Y = G(X) + H(X) dY/dX = dG/dX + dH/dX

Functions

example Y = 5•X + 5•X2 dY/dX = 5 + 10•X

• Product of Y = G(X)•H(X)

Two Functions 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

Quick Differentiation Review

Slide 6

• Quotient of Two Y = G(X) / H(X) Functions

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

Quick Differentiation Review

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 - Q2, then d/dQ = 50 - 2·Q, using the rules of differentiation.

• Hence, Q = 25 will maximize profits where 50 - 2•Q = 0.

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·Q2 - 60·Q, then dC/dQ = 10·Q - 60.

• Hence, Q = 6 will minimize cost where 10•Q - 60 = 0.

Slide 9

More Examples

• Competitive Firm: Maximize Profits » where = TR - TC = P•Q - TC(Q)» Use our first order condition:

d/dQ = P - dTC/dQ = 0. » Decision Rule: P = MC.

a function of Q

Max = 100•Q - Q2

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

Max= 50 + 5•X2

So, 10•X = 0 implies Q = 0 and= 50

Problem 1 Problem 2

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 = 100•Q - Q2

100 -2•Q = 0

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

Max= 50 + 5•X2

10•X = 0

second derivative is: 10 implies Q = 0 is a MIN

Problem 1 Problem 2

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.

Slide 12

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 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 -10X2 -20Y2 +20XY = 3,250