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is positive, the point is a local minimum. Taking the derivative of d/dQ, we

find the second derivative to be

In finding the second derivative, we start from the original profit function and

take the derivative twice.At Q 2, we find that d^2 /dQ^2 3.6  .6(2) 2.4.

Since this is positive, Q 2 represents a local minimum. At Q 10, we find that

d^2 /dQ^2 3.6  .6(10) 2.4. Since this is negative, Q 10 represents a

local maximum.

MARGINAL REVENUE AND MARGINAL COST We have seen that maximum

profit is achieved at the point such that marginal profit equals zero, d/dQ 0.

The same condition can be expressed in a different form by separating profit

into its two components. Profit is defined as the difference between revenues

and costs. Thus, the profit function can be written as

the difference between revenues and costs. In turn, the condition that mar-

ginal profit equal zero is

In short, profit is maximized when marginal revenue equals marginal cost.

Maximizing Multivariable Functions

Frequently, the manager must determine optimal values for several decision

variables at once, for instance, a product’s price and its associated advertising

budget. In this case, the product’s profit would be expressed by the function,

(P, A), where P is the product’s price and A is its advertising budget in

dollars. Here the key to maximizing profit is to apply a double dose of mar-

ginal reasoning. Marginal profit with respect to each decision variable should

be equated to zero. The optimal value of P is found where the “partial” deriv-

ative of profit with respect to P equals zero. This partial derivative is denoted

by 

/ P and is found by taking the derivative with respect to P, holding A (the

other decision variable) constant. Similarly, the optimal value of A is found

where 

/ A 0.

d/dQdR/dQdC/dQMRMC0.

(Q)R(Q)C(Q),

3.6.6Q

dM/dQd(3.6Q.3Q^2 6)/dQ

d^2 /dQ^2 d(d/dQ)/dQ

68 Appendix to Chapter 2 Calculus and Optimization Techniques

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