Chapter 5.4, Problem 3E

To determine

Find the smallest perimeter possible for a rectangle.

Answer to Problem 3E

the minimum perimeter the dimensions are $\left(4,4\right)$ in and perimeter is $16in$

Explanation of Solution

Given information: Given area of the rectangle is $16c{m}^2$ .

Calculation:

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

  $\begin{array}{l}\text{then}\\ A=xy=16\\ y=\frac{16}{x}\end{array}$

Let the perimeter of the rectangle be

  $P\text{ , }0<x<4$

  $\begin{array}{l}P=2\left(x+y\right)\\ =2\left(x+\frac{16}{x}\right)\end{array}$

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

  $\begin{array}{l}P=2\left(x+\frac{16}{x}\right)\\ \frac{dP}{dx}=2-\frac{32}{x^3}\end{array}$

Critical points occur at

  $\begin{array}{l}{P}^{\prime }=0\\ 2-\frac{32}{x^3}=0\\ x=\pm \sqrt{16}\\ =\pm 4\end{array}$

Consider $x=4\text{ since }x>0$ ,solve it graphically, it has critical point occur at $x=4$

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

For

  $\begin{array}{l}x=4\\ y=\frac{16}{x}\\ =4\\ \text{ P}=2\left(x+\frac{16}{x}\right)\\ =2\left(4+4\right)\\ =16\end{array}$

Graphical support:

  

Thus, the minimum perimeter the dimensions are $\left(4,4\right)$ in and perimeter is $16in$