Chapter 4, Problem 50RE

(a)

To determine

To find: The equation of the volume.

Answer to Problem 50RE

The required expression for volume is $V=2x^3-25x^2+75x$ .

Explanation of Solution

Given information:

The piece of cardboard is $10\times 15$ inches. Two equal square are removed from the corner of the side 15 inches.

Calculation:

  

The height of the prism can be x, which is the labeled length of the rectangle next to the base and lid.

In order to write a formula in terms of the variable x, it is required to write expressions for the length, width, and height of the box.

It can be said that the length is the bottom of the base the prism. To find this value, we can subtract $x$ and $x$ from 10 and divide this by 2, since there are two equal rectangles (base and lid).

  $\begin{array}{l}length=\frac{10-2x}{2}\\ l=5-x\end{array}$

The width is the side of the base of the prism. To find this value, we can subtract $2x$ from 10.

  $width=15-2x$

The volume of the rectangular box is given by:

  $V=l\times w\times h$

Substitute the value of length width and height gives:

  $\begin{array}{l}V=x\left(15-2x\right)\left(5-x\right)\\ V=2x^3-25x^2+75x\end{array}$

This is the required expression for volume.

(b)

To determine

To find: The domain of $V$ .

Answer to Problem 50RE

The required domain is $\left(0,5\right)$ .

Explanation of Solution

Given information:

The required expression for volume is $V=2x^3-25x^2+75x$ .

Calculation:

Let the height of the prism be $x$ , which is the labeled length of the rectangle next to the base and lid.

  

In order to write a formula in terms of the variable $x$ , it is required to write expressions for the length, width, and height of the box.

It can be said that the length is the bottom of the base of the prism. To find this value, we can subtract x and x from 10 and divide this by 2, since there are two equal rectangles (base and lid).

  $\begin{array}{l}length=\frac{10-2x}{2}\\ l=5-x\end{array}$

The width is the side of the base of the prism. To find this value, we can subtract $\text{2x}$ from 15.

  $width=15-2x$

The domain of the volume is where $\text{x > 0}$ , the length width and height cannot be negative therefore $,15-2x>0,10-2x>0,x>0$ .

Combining these requirements the domain is $\left(0,5\right)$ .

(c)

To determine

To find: Use the graphical method to find the maximum value of $V$ and where it is found.

Answer to Problem 50RE

It is observed from the graph that the maximum volume is 66.02 is at $x=1`.96$

Explanation of Solution

Given information:

The volume function is $V=2x^3-25x^2+75x$

Calculation:

In order to draw the graph of volume function substitute different values of $x$ and find the corresponding values for $V$ .

Let $x=1.2$ then

  $\begin{array}{l}V=2{\left(1.2\right)}^3-25{\left(1.2\right)}^2+75\cdot 1.2\\ V=2\cdot 1.728-25\cdot 1.44+450\\ V=3.456-46+90\\ V=57.69\end{array}$

The point will be $\left(1.2,57.69\right)$ .

Similarly, more point can be found and plot the on the graph. The graph is given below:

  

It is observed from the graph that the maximum volume is 66.02 is at $x=1`.96$

(c)

To determine

To find: confirm your result in part (c) analytically.

Answer to Problem 50RE

It is found analytically that the maximum volume is 66.02 is at $x=1`.96$

Explanation of Solution

Given information:

Volume function is $V=2x^3-25x^2+75x$

Calculation:

In order to solve the equation $V=2x^3-25x^2+75x$ it can be rewritten as:

  $V=x(2x^2-25x+75)$

In order to find the maximum value, it is required to find the critical points.

In order to find critical points differentiate volume function with respect to $x$ and equate it to zero.

  $\begin{array}{l}\frac{d}{dx}\left(V\right)=\frac{d}{dx}\left(2x^3-25x^2+75\right)\\ V\text{'}=\frac{d}{dx}\left(2x^3\right)-\frac{d}{dx}\left(25x^2\right)+\frac{d}{dx}\left(25\right)\\ V\text{'}=\left(6x^2-50x+75\right)\\ V\text{'}=0\end{array}$

Now substitute $V\text{'}=0$ :

  $6x^2-50x+75=0$

Use quadratic formula $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ to solve this quadratic equation and find critical points.

  $\begin{array}{l}x=\frac{-\left(-50\right)\pm \sqrt{{\left(-50\right)}^2-4\cdot 6\cdot 75}}{2\cdot 6}\\ x=\frac{50\pm \sqrt{2500-1800}}{12}\\ x=\frac{50\pm \sqrt{700}}{12}\\ x=\frac{50\pm 10\sqrt{7}}{12}\\ x=\frac{25\pm 5\sqrt{7}}{6}\\ x=1.96\\ x=6.39\end{array}$

Discard the largest value because it is not in the domain $\left(0,5\right)$ .

Perform second order derivative test to check whether the volume is maximum.

Now differentiate again $V\text{'}=6x^2-50x+75$ .

  $\begin{array}{l}V\text{'}\text{'}=12x-50\\ V\text{'}\text{'}=12\left(1.96\right)-50\\ V\text{'}\text{'}=-23.96\end{array}$

  $V\text{'}\text{'}$ is negative at $x=1.96$ .

  $V\text{'}\text{'}=12\left(1.96\right)-50<0$

Now find maximum value of volume by substituting $x=1.96$ in $V=2x^3-25x^2+75x$

  $\begin{array}{l}{V}_{\max }=2{\left(1.96\right)}^3-25{\left(1.96\right)}^2+75\left(1.96\right)\\ V_{\max }=66.02\end{array}$

This proves that the volume is maximum at $x=1.06$