Chapter 5.4, Problem 4E

To determine

Prove among all rectangle with an 8-m perimeter .

Answer to Problem 4E

the maximum area of the sides are $\left(2m,2m\right)$ which is square and area is $4m^2$

Explanation of Solution

Given information: Given perimeter of the rectangle is $8m$ .

Calculation:

Let the sides are $x\text{ and }y$

  $\begin{array}{l}\text{then}\\ 2\left(x+y\right)=8\\ \left(x+y\right)=4\\ y=4-x\end{array}$

Let the area of the rectangle be

  $\text{A , }0<x<4$

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

For any function $f\left(x\right)$ the condition of minimum and maximum is obtained by find $f^{\prime }\left(x\right)=0$ ,

Critical points and end points

  $\begin{array}{l}A=4x-x^2\\ \frac{dA}{dx}=4-2x\end{array}$

Critical points occur at

  $\begin{array}{l}{A}^{\prime }=0\\ 4-2x=0\\ x=2\end{array}$

Consider $x=2$ ,solve it graphically, it has critical point occur at $x=2$

  $\frac{dA}{dx}>0\text{ for 0< }x<2\text{and}\frac{dA}{dx}<0\text{ for2< }x<4$ this corresponds to maximum area.

For

  $\begin{array}{l}x=2\\ y=4-x\\ =4-2\\ =2\\ \text{ A}=\left(x\right)\left(y\right)\\ =2\left(2\right)\\ =4\end{array}$

Graphical support:

  

Thus, the maximum area of the sides is $\left(2m,2m\right)$ which is square and area is $4m^2$