Chapter 2.7, Problem 35E

Solution

a.

Express the perimeter P as a function of x.

  $P=2x+\frac{364}{x}$

Given:

Rectangles with an area of $182{\text{ ft}}^2$ . Let x be the length of one side of such rectangle.

Calculation:

Area of rectangle is given by $182{\text{ ft}}^2$ .

Let the width of the rectangle is y. then $y=\frac{182}{x}$ .

Now, perimeter of the rectangle is given by

  $\begin{array}{c}P=2\left(x+y\right)\\ =2\left(x+\frac{182}{x}\right)\\ =2x+\frac{364}{x}\end{array}$

Therefore, the perimeter P as a function of x is given by

  $P=2x+\frac{364}{x}$

b.

Find the dimension of the rectangle that has the least perimeter. Find the least perimeter as well.

  $x=13.49\text{ ft}$ .

  $P=53.96\text{ ft}$

Given:

  $P=2x+\frac{364}{x}$

Length is x and width is $\frac{182}{x}$ .

Calculation:

It has been given that $P=2x+\frac{364}{x}$ .

Differentiate with respect to x

  $P\text{'}=2-\frac{364}{x^2}$

To find the critical points

  $\begin{array}{c}P\text{'}=0\\ 2-\frac{364}{x^2}=0\\ \frac{364}{x^2}=2\\ x^2=182\\ x=13.49\end{array}$

Hence, for the minimum P, the length is given by $x=13.49\text{ ft}$ .

And the width of the rectangle is given by $\frac{182}{x}=\frac{182}{13.49}=13.49\text{ ft}$ .

The least perimeter is given by

  $\begin{array}{c}P=2x+\frac{364}{x}\\ =2\cdot 13.49+\frac{364}{13.49}\\ =53.96\text{ ft}\end{array}$