Chapter 1, Problem 68RE

Solution

(a).

To write: The area $A$ of the rectangle as a function of $x$ .

The area $A$ of the rectangle as a function of $x$ is equal to $72x-2x^3$ .

Given information:

A rectangle is inscribed between the $x-$ axis and the parabola $y=36-x^2$ with one side along the $x-$ axis, as shown in the figure below:

  

Calculation:

Due to symmetry, the length of the rectangle is $x+x=2x$ and the width is $y$ .

Formula used:

The area of the rectangle is given by

  $A=\text{length}\times \text{width}$

Apply the above formula and calculate the area of the rectangle can be expressed below:

  $\begin{array}{c}A=2x\times y\\ =2xy\end{array}$

Substitute $y=36-x^2$ in the above equation:

  $\begin{array}{c}A=2x\left(36-x^2\right)\\ =72x-2x^3\end{array}$

Therefore, the area $A$ of the rectangle as a function of $x$ is $72x-2x^3$ .

(b).

To find: The value of $x$ are in the domain of $A$ .

The values of $x$ are belong to $\left[0,6\right]$ .

Given information:

A rectangle is inscribed between the $x-$ axis and the parabola $y=36-x^2$ with one side along the $x-$ axis, as shown in the figure below:

  

Calculation:

From the part (a),

  $A=72x-2x^3$

For the domain of the area, substitute $36-x^2\ge 0$ and solve for $x$ :

  $\begin{array}{c}72x-2x^3\ge 0\\ 2x\left(36-x^2\right)\ge 0\\ 2x\ge 0\text{or}36-x^2\ge 0\\ x\ge 0\text{or}{x}^2\le 36\\ x\ge 0\text{or}x\le \pm 6\\ x\in \left[0,6\right]\end{array}$

Therefore, the values of $x$ are belong to $\left[0,6\right]$ .

(c).

To sketch: The graph area $A$ over the domain.

The graph of the area $A$ is below:

  

Given information:

A rectangle is inscribed between the $x-$ axis and the parabola $y=36-x^2$ with one side along the $x-$ axis, as shown in the figure below:

  

Calculation:

From the part (a) and part (b),

  $A=72x-2x^3$

  $x\in \left[0,6\right]$

Now, draw the graph of the area $A$:

  

(d).

To find: The maximum area from the graph.

The maximum area is equal to $166.277$ .

Given information:

A rectangle is inscribed between the $x-$ axis and the parabola $y=36-x^2$ with one side along the $x-$ axis, as shown in the figure below:

  

Calculation:

From the part (c),

The graph of the area is below:

  

From the above graph, the maximum area is $166.277$ .

Therefore, the maximum area is $166.277$ .