Chapter 4.4, Problem 4E

To determine

To prove: To show that among all rectangles with an 8m perimeter, the one with largest area is a square.

Explanation of Solution

Formula used:

Perimeter of the rectangle, $P=2\left(l+w\right)$

Area of the rectangle, $A=l\cdot w$

Where $l$ is the length and $w$ is the width.

Proof: Given that the perimeter of the rectangle is 8m.

That is,

  $\begin{array}{l}P=2\left(l+w\right)\\ P=8\end{array}$

By solving the above equation,

  $\begin{array}{l}l+w=4\\ w=4-l\end{array}$

Substituting this equation to calculate the area of rectangle,

  $\begin{array}{l}A=l\cdot w\\ =l\left(4-l\right)\\ =4l-l^2\end{array}$

Upon taking first derivative of the above area equation,

  $\frac{dA}{dl}=4-2l$

Assuming that, $\frac{dA}{dt}=0$

That is, $4-2l=0$

So,

  $\begin{array}{l}2l=4\\ l=2\end{array}$

Upon substituting the value of l in the equation, $w=4-l$ , the value of was 2.

Thus the dimensions of the rectangle are, $l=2$ and $w=2$ . Therefore, it must be a square.