Chapter 4.4, Problem 6E

To determine

To find: To find the area of the rectangle.

Answer to Problem 6E

The area of the rectangle is $32$ .

Explanation of Solution

Given:

The Area of the rectangle is $A=2x(12-x^2)$ .

Calculation:

Given the vertices pf the rectangle are on the parabola $y=12-x^2$ .

The upper vertices of the rectangle are $x,12-x^2$ and $0<x<\sqrt{12}$ .

Then the dimensions of the rectangle are $2x,12-x^2$

The area of the rectangle be A.

  $A=2x(12-x^2)$

For any function $f(x)$ the condition of minimum and maximum is obtained by finding

  $f\text{'}(x)=0$ , the critical points and end points.

  $\begin{array}{l}A=2x(12-x^2)\\ \frac{dA}{dx}=24-6x^2\end{array}$

Critical points occur at,

  $\begin{array}{l}A\text{'}=0\\ 24-6x^2=0\\ x=2\end{array}$

Solving it graphically, we get it has critical point occur at $x=2$ .

  $\frac{dA}{dx}>0$ for $0<x<2$ and $\frac{dA}{dx}<0$ for $2<x<\sqrt{12}$ this corresponds to maximum area.

For,

  $\begin{array}{l}x=2\\ y=12-x^2\\ y=12-4\\ y=8\\ A=(2x)(12-x^2)\\ A=32\end{array}$