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746 MATHEMATICAL APPENDIX

3. How much will the area change when the amount of where is the Lagrange multiplier. Note that we have
2
fence varies? We know that the Area LW F /16. rewritten the constraint so that the right-hand side is zero
Therefore, dArea/dF F/8. The area will increase by (i.e., 10 T R 0). We then place the left-hand side of
about F/8 square feet when the perimeter is increased the constraint in the Lagrangian function.
by one foot. The conditions for an interior optimum (with T 0
and R 0) are

A.8 LAGRANGE MULTIPLIERS
0¶ 0B(T, R)
0 ¡ l 0 (A.19)
In the previous section we showed how to solve a con- 0T 0T
strained optimization problem by solving the constraint for
one of the variables and then substituting the constraint 0¶ 0 ¡ 0B(T, R) l 0 (A.20)
into the objective function. This technique is most likely 0R 0R
to work when the constraint (or set of constraints) has a
0¶
simple form. However, it may not be possible to use this 0 ¡ 10 T R 0 (A.21)
approach in more complicated problems. 0l
We now show how to solve constrained optimization
problems by constructing an equation, called the Lagrangian The partial derivatives in this problem are B(T, R)/
0
function, that is a combination of the objective function and 0 T 5000 500T and B(T, R)/ R 1000 100R. Thus,
0
0
the constraint. We begin with a general description of the we can write (A.18) as
method, and then illustrate how to use it with two Learning-
By-Doing Exercises. 5000 500T l, and (A.22)
We first construct the Lagrangian function as follows:

(x, y, ) F(x, y) G(x, y) . This function is the sum of 1000 100R l (A.23)
two terms: (1) the objective function, and (2) the constraint,
multiplied by an unknown factor, , which is called the Since the right-hand sides of equations (A.22) and
Lagrange multiplier. We then set the partial derivatives of the (A.23) are the same (l) , we know that at an optimum
Lagrangian function with respect to the three unknowns 5000 500T 1000 100R. This is equation (A.24).
(x, y, and ) equal to zero. Equation (A.25) is the same as equation (A.21). Together,
equations (A.24) and (A.25) give us two equations in two
0¶ 0F(x, y) 0G(x, y) unknowns, T and R. We now know that the optimal
0 ¡ l 0 (A.16)
0x 0x 0x amounts of radio and television advertising are determined
by two equations:
0¶ 0F(x, y) 0G(x, y)
0 ¡ l 0 (A.17)
0y 0y 0y 5000 500T 1000 100R (A.24)
0¶ T R 10 (A.25)
0 ¡ G(x, y) 0 (A.18)
0l
We then find that T $8.33 (hundred thousand) and
We can then use the three equations (A.16, A.17, and R $1.67 (hundred thousand), the same solution we found
A.18) to solve for the three unknowns. To see how to apply in Learning-By-Doing Exercise A.9.
this method, consider the following two exercises. It is also possible to calculate the value of the Lagrange
l
multiplier at the optimum, and this value has an impor-
tant economic interpretation. We observe that l 5000
LEARNING-BY-DOING EXERCISE A.11
500T 5000 500(25/3) 833.33 . (Alternatively, l
Radio and Beer Advertising Revisited 1000 100R 1000 100(5/3) 833.33.) The value of
l tells us (approximately) how much beer sales (the objec-
Problem The problem here is the same as in Learning- tive function) could be increased if the advertising budget
By-Doing Exercise A.9. Now let’s solve the problem using were increased by one “unit” (in this problem a unit of ad-
the method of Lagrange multipliers. vertising is $100,000). The manager could expect sales to
increase by about 833 barrels for every $100,000 in extra
Solution We define the Lagrangian function
advertising, or by about 0.00833 barrels for each additional
dollar of advertising.
¶(T, R, l) B(T, R) l(10 T R)