Chapter 5.4, Problem 6E

To determine

Find the largest area of the rectangle.

Answer to Problem 6E

The maximum area of the sides is $\left(4,8\right)$ which is square and area is $32$

Explanation of Solution

Given information: Given parabola is $y=12-x^2$

Calculation:

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

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

Let the area of the rectangle be

  $\text{A , }$

  $\begin{array}{l}A=2x\left(12-x^2\right)\\ =24x-2x^3\\ =4x-x^2\end{array}$

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

  $\begin{array}{l}A=24x-2x^3\\ \frac{dA}{dx}=24-6x^2\end{array}$

Critical points occur at

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

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

Here $\frac{dA}{dx}>0\text{ for 0< }x<2\text{and}\frac{dA}{dx}<0\text{ for2< }x<\sqrt{12}$ this corresponds to maximum area.

For

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

Graphical support:

  

Thus, the maximum area of the sides is $\left(4,8\right)$ which is square and area is $32$ .