Suppose a farmer had a certain length of fence, P, and wished to enclose the largest possible rectangular area. What shape area should the farmer choose? This is clearly a problem in constrained maximization. To solve it, let x be the length of one side of the rectangle and y be the length of the other side. The problem then is to choose x and y so as to maximize the area of the field (given by A = x ⋅y), subject to the constraint that the perimeter is fixed at

P = 2x + 2y.

Setting up the Lagrangian expression gives

 

 

The three equations in 2.63 must be solved simultaneously for x, y, and λ. The first two equations say that y/2 = x/2 = λ, showing that x must be equal to y (the field should be square). They also imply that x and y should be chosen so that the ratio of marginal benefits to marginal cost is the same for both variables. The benefit (in terms of area) of one more unit of x is given by y (area is increased by 1 .y), and the marginal cost (in terms of perimeter) is 2 (the available perimeter is reduced by 2 for each unit that the length of side x is increased). The maximum conditions state that this ratio should be equal for each of the variables. Since we have shown that x = y, we can use the constraint to show that

 

Interpretation of the Lagrangian Multiplier.If the farmer were interested in knowing how much more field could be fenced by adding an extra yard of fence, the Lagrangian multiplier suggests that he or she could find out by dividing the present perimeter by 8. Some specific numbers might make this clear. Suppose that the field currently has a perimeter of 400 yards. If the farmer has planned “optimally,” the field will be a square with 100 yards (= P/4) on a side. The enclosed area will be 10,000 square yards. Suppose now that the perimeter (that is, the available fence) were enlarged by one yard. Equation 2.65 would then “predict” that the total area would be increased by approximately 50 (= P/8) square yards. That this is indeed the case can be shown as follows: Because the perimeter is now 401 yards, each side of the square will be 401/4 yards. The total area of the field is therefore (401/4)², which, according to the author’s calculator, works out to be 10,050.06 square yards. Hence, the “prediction” of a 50- square-yard increase that is provided by the Lagrangian multiplier proves to be remarkably close. As in all constrained maximization problems, here the Lagrangian multiplier provides useful information about the implicit value of the constraint.

Duality.The dual of this constrained maximization problem is that for a given area of a rectangular field, the farmer wishes to minimize the fence required to surround it. Mathematically, the problem is to minimize

 

(where the D denotes the dual concept) yields the following first-order conditions for a minimum:

Solving these equations as before yields the result

Again, the field should be square if the length of fence is to be minimized. The value of the Lagrangian multiplier in this problem is

As before, this Lagrangian multiplier indicates the relationship between the objective (minimizing fence) and the constraint (needing to surround the field). If the field were 10,000 square yards, as we saw before, 400 yards of fence would be needed. Increasing the field by one square yard would require about .02 more yards of fence (= 2/ √A = 2/100).The reader may wish to fire up his or her calculator to show this is indeed the case—a fence 100.005 yards on each side will exactly enclose 10,001 square yards. Here, as in most duality problems, the value of the Lagrangian in the dual is the reciprocal of the value for the Lagrangian in the primal problem. Both provide the same information, although in a somewhat different form.