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Example 2: Fencing the Maximum Area

A farmer plans to use 1000 feet of fence to enclose a rectangular plot along the bank of a straight river. Find the dimensions which enclose the maximum area.

03_continuous_functions-213.gif

Figure 3.6.2

Let x be the dimension of the side along the river, and y be the other dimension, as in Figure 3.6.2. Call the area A.

No fencing is needed on the side of the plot bordering the river. The given information is expressed by the following system of formulas.

A = xy, x + 2y = 1000, 0 ≤ x ≤ 1000.

The problem is to find the values of x and y at which A is maximum. In this problem A is expressed in terms of two variables instead of one. However, we can select x as the independent variable, and then both y and A are functions of x. We find an equation for A as a function of x alone by eliminating y.

x + 2y = 1000, 03_continuous_functions-214.gif

03_continuous_functions-215.gif

We then find the maximum of A in the closed interval 0 ≤ x ≤ 1000.

Step 1

dA/dx = 500 - x.

Step 2

dA/dx = 0 when x = 500. This is the unique interior critical point.

Step 3

We use the Second Derivative Test:

d2A/dx2 = -1.

Therefore A has a maximum at the critical point x = 500.

CONCLUSION

The maximum area occurs when the plot has dimensions x = 500 ft and y = (1000 - x)/2 = 250ft (Figure 3.6.3).

03_continuous_functions-216.gif

Figure 3.6.3

Home Continuous Functions Maxima and Minima - Applications Examples Example 2: Fencing the Maximum Area

Last Update: 2006-11-25