Question

Let x be the length (in feet) of each of the two sides perpendicular to the barn wall, and let y be the length of the side parallel to the wall. The farmer has 240 ft of fencing to use on the two perpendicular sides and the one parallel side (the wall itself needs no fencing). 1. Write the constraint relating x and y based on the total fencing available. 2. Use this constraint to express the enclosed area A as a function of x alone. 3. Determine the value of x that maximizes the area, and find the maximum possible area. What is the maximum area (in square feet) that can be enclosed?

Accepted Answer

Since fencing is used on the two sides of length $x$ and the one side of length $y$ , the constraint is $2x + y = 240$ , so $y = 240 - 2x$ . Substituting into $A = xy$ gives $A(x) = x(240-2x) = 240x - 2x^2$ , defined for $0 < x < 120$ so that both $x$ and $y$ stay positive.

Taking the derivative, $A'(x) = 240 - 4x$ . Setting $A'(x) = 0$ gives $x = 60$ . Since $A''(x) = -4 < 0$ , this critical point is a maximum. Then $y = 240 - 2(60) = 120$ , so the maximum area is $A(60) = 60 \cdot 120 = 7200$ square feet.

Maximizing area of a three-sided fenced rectangle

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