Fencing an Enclosure with an Internal Divider
A farmer has 180 m of fencing to make a rectangular enclosure against a straight river, using the river as one side. The enclosure is split into two equal pens by a fence perpendicular to the river, so each pen has river frontage. Let x be the total length along the river and y the depth. A worker says the maximum area must occur when the enclosure is a square, because squares maximize area for a fixed perimeter. Another worker says the extra divider changes the problem, so the square intuition does not apply. The farmer also has a property line that limits the river frontage x to at most 80 m.
1. Write the fencing constraint and area function, find the unconstrained dimensions that maximize area, and compare with the square case. Which worker is right?
Read the worked answer
The fencing consists of one back side of length x, two side fences of length y, and one divider of length y. So x + 3y = 180, giving y = (180 − x)/3. The area is A(x) = xy = x(180 − x)/3 = 60x − x²/3, with 0 < x < 180.
Differentiating: A′(x) = 60 − 2x/3. Setting A′(x) = 0 gives 60 = 2x/3, so x = 90. Then y = (180 − 90)/3 = 30, and the maximum area is 90 × 30 = 2700 m².
If the enclosure were a square, x = y. Then x + 3x = 180, so x = 45 and the area is 2025 m². The second worker is right: the divider changes the fencing equation, and the optimum is 90 m by 30 m, not square.
2. With the property line x ≤ 80, find the constrained maximum and explain why the derivative test alone is insufficient.
Read the worked answer
On 0 < x < 90, A′(x) = 60 − 2x/3 is positive, so the area is increasing up to the unconstrained optimum at x = 90. Since the property line requires x ≤ 80, the critical point at 90 is outside the feasible interval. The maximum on the interval must occur at the boundary x = 80.
At x = 80, y = (180 − 80)/3 = 100/3 ≈ 33.33 m, and A = 80 × 100/3 = 8000/3 ≈ 2666.67 m². The derivative test alone is insufficient because the critical point is not feasible; one must compare boundary values, and x = 0 gives area 0. Thus the constrained maximum is about 2666.67 m² at x = 80 m.