Home › Geometry and measurement

Maximize a rectangular area with fencing on three sides

Count only the fenced sides

For a rectangle beside a river, two widths $w$ and one length $l$ use the fence: $2 w + l = F$ . The area is $A (w) = w (F - 2 w)$ , a downward-opening quadratic with its maximum at $w = F /4$ .

Problem

A farmer has 5,500 meters of fencing for a rectangular plot beside a straight river. The side along the river needs no fence. What is the greatest area that can be enclosed?

Answer

3, 781, 250

square meters.

At the maximum, the two widths are 1,375 meters each and the fenced length is 2,750 meters.

Step-by-step solution

1. Express the length using the fence

Let each side perpendicular to the river have width $w$ . Then the third fenced side has length $5500 - 2 w$ , with $0 < w < 2750$ .

2. Make an area function

The area is $A (w) = w (5500 - 2 w) = 5500 w - 2 w^{2}$ . Since the coefficient of $w^{2}$ is negative, the vertex gives the maximum.

3. Find and check the maximum

The vertex occurs at $w = - 5500/ (2 (- 2)) = 1375$ . The third side is $5500 - 2 (1375) = 2750$ , so the area is $1375 (2750) = 3, 781, 250$ square meters.

Add to Chrome - solve your own

Screenshot any problem on your screen and get the steps. Chrome on a computer; free.