Chapter 4.1, Problem 44E

(a)

To determine

To find : extreme values of $P$ .

Answer to Problem 44E

Minimum occur at $\left(10,40\right)$ .

Explanation of Solution

Given information :

The perimeter of a rectangle is $P\left(x\right)=2x+\frac{200}{x},0<x<\infty$ .

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

Differentiate $P\left(x\right)=2x+\frac{200}{x}$ with respect to $x$ .

  $\begin{array}{c}{P}^{\prime }\left(x\right)=\frac{d}{dx}\left(2x+\frac{200}{x}\right)\\ =2-\frac{200}{x^2}\\ =\frac{2x^2-200}{x^2}\\ =\frac{2\left(x^2-100\right)}{x^2}\end{array}$

Take $P^{\prime }\left(x\right)=0$ for critical points.

  $\begin{array}{c}\frac{2\left(x^2-100\right)}{x^2}=0\\ \frac{\left(x-10\right)\left(x+10\right)}{x^2}=0\left[2\ne 0\right]\\ \left(x-10\right)\left(x+10\right)=0\left[x^2\ne 0\right]\\ x=-10,10\end{array}$

Only one critical point exists in the given interval that is $x=10$ .

Plug the critical point in the function.

  $\begin{array}{c}P\left(10\right)=2\cdot 10+\frac{200}{10}\\ =20+20\\ =40\end{array}$

As the $x=40$ move away from both ends, the value of P increases.

Hence, the function has minimum at $\left(10,40\right)$ .

(b)

To determine

To interpret : any values found in part (a).

Explanation of Solution

Given information :

The perimeter of a rectangle is $P\left(x\right)=2x+\frac{200}{x},0<x<\infty$ .

Formula needed :

For critical points, $\frac{dy}{dx}=0$ or $\frac{dy}{dx}$ is not defined.

The function has minimum at $\left(10,40\right)$ . This means the smallest perimeter of rectangle is 40 when side of rectangle is $x=10$ .

Also, the rectangle becomes square when $x=10$ .