Question

Prove that the most efficient way to build a rectangular fenced area along a river-so that only three sides of fencing are needed-is to make the side parallel to the river twice as long as the other sides. You may assume that you have a fixed amount of fencing material.

Short answer

The area is maximized when $x=\frac{P}{4}$, it implies that when the rectangle is a square.

Step-by-step solution

Given information.

The goal is to demonstrate that, given a specified perimeter P, the rectangle with the largest feasible area is always a square.

Explanation of the statement.

Consider, $x$and $y$ are the sides of the rectangle.

The rectangle's perimeter is then $P=2x+2y.$

So,

$y=\frac{1}{2}(P-2x)$

The area is as follows.

$$\begin{gathered} A=xy \\ =x\left(\frac{1}{2}(P-2x)\right) \end{gathered}$$

The area's derivation is as follows:

$$\begin{gathered} A=x\left(\frac{1}{2}(P-2x)\right) \\ {A}^{\text{'}}=\frac{1}{2}P-2x=0 \\ x=\frac{P}{4} \end{gathered}$$

The only critical point is $x=\frac{P}{4}$.

since,$A^{\text{'}}\left(\frac{P}{4}-1\right)>0$and $A^{\text{'}}\left(\frac{P}{4}+1\right)<0.$

The first derivative so indicates that $A\left(x\right)$consists of a local maximum at $x=\frac{P}{4}.$

The highest possible value is at $\left[0,\frac{P}{2}\right]$.

The area is therefore maximised when $x=\frac{P}{4}$, i.e if the rectangle is square in shape.