Question

Prove that the rectangle with the largest possible area given a fixed perimeter P is always a square.

Short answer

The rectangle with the largest possible area given a fixed perimeter P is always a square has been proved.

Step-by-step solution

Step 1. Given information.

We have to prove that the rectangle with the largest possible area given a fixed perimeter P is always a square.

Step 2. Find the area.

Let x and y be the sides of the rectangle.

Thus,

$P=2x+2y$

and,

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

The area is :

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

Step 3. Derivative of the area.

Derivative of the area is:

$\begin{array}{r}A=x\left(\frac{1}{2}(P-2x)\right)\\ A^{\mathrm{\prime }}=\frac{1}{2}P-2x=0\\ x=\frac{P}{4}\end{array}$

Critical point is $x=\frac{P}{4}$

Since,

$A^{\mathrm{\prime }}\left(\frac{P}{4}-1\right)>0\text{and}{A}^{\mathrm{\prime }}\left(\frac{P}{4}+1\right)<0$

So, the first derivative gives us the local maximum at $x=\frac{P}{4}$

Global maximum =$\left[0,\frac{P}{2}\right]$