Chapter 4.6, Problem 10E

(a)

To determine

To estimate: The volume of configurations by plotting several diagrams of the open top box with different bases and height.

Answer to Problem 10E

The maximum volume is ${\text{2ft}}^3$.

Explanation of Solution

Given: The width of the cardboard $=3$ ft

To draw:

Several diagrams illustrating the situation with different bases and height of the open top box.

Case 1: Draw an open top box with base length 1 ft and width 1 ft and height 1 ft.

In the Figure 1,

The width of the cardboard is $1+1+1=3\text{ft}$

The volume of the open top box is 1 ${\text{ft}}^3$

Case 2: Draw an open top box with base length $\frac{3}{2}$ ft and width $\frac{3}{2}$ ft and height $\frac{3}{4}$ ft.

In the Figure 2,

The width of the cardboard is $\frac{3}{4}+\frac{3}{4}+\frac{3}{2}=3\text{ft}$

The volume of the open top box is $\left(\frac{3}{4}\times \frac{3}{2}\times \frac{3}{2}\right)\text{=1}{\text{.6875ft}}^3$

Case 3: Draw an open top box with base length 2 ft and width 2 ft and height $\frac{1}{2}$ ft

In the Figure 2,

The width of the cardboard is $2+\frac{1}{2}+\frac{1}{2}=3\text{ft}$

The volume of the open top box is $\left(2\times 2\times \frac{1}{2}\right){\text{=2ft}}^3$

By comparing, it is clear that, among all the 3 cases, the maximum volume is ${\text{2ft}}^3$.

(b)

To determine

To draw: The diagram illustrating the general solution for the open top box with different bases and height.

Explanation of Solution

Given:

A box with an open top is to be constructed from a square piece of cardboard, 3 ft wide, by cutting out a square from each of the four corners and bending up the sides.

Draw an open top box with length y and width y and height x.

In Figure 4,

The width of the cardboard is $x+y+x=2x+y$

(c)

To determine

To find: the volume of the box with length y and width y and height x.

Answer to Problem 10E

The volume of the open top box is $x{y}^2$ ${\text{ft}}^3$

Explanation of Solution

Volume $V=$ Length $\times$ width $\times$ height

$\begin{array}{c}V=y\times y\times x\\ =x{y}^2\end{array}$

The volume of the open top box is $x{y}^2$ ${\text{ft}}^3$

(d)

To determine

To find: The equation of the width of the cardboard.

Answer to Problem 10E

The equation is $y+2x=3$

Explanation of Solution

Given:

The cardboard is 3 ft wide.

Length of cardboard $=3$

Calculation:

In Figure 5,

$x+y+x=3$

$y+2x=3$

The relatable equation is $y+2x=3$

(e)

To determine

To find: The volume of the open top box in one variable.

Answer to Problem 10E

The volume is $v=x{\left(3-2x\right)}^2$.

Explanation of Solution

Given:

The volume of the open top box is $x{y}^2$ ${\text{ft}}^3$.

Calculation:

In Figure 6,

$\begin{array}{l}y+3x=3\\ y=3-2x\\ \end{array}$

Substitute the value of $y=3-2x$ in $V=x{y}^2$

$v=x{\left(3-2x\right)}^2$

Thus, the volume is $v=x{\left(3-2x\right)}^2$.

(f)

To determine

To solve: The problem using calculus and to compare the answer with the estimated volume ${\text{2ft}}^3$.

Answer to Problem 10E

The maximum volume is ${\text{2ft}}^3$.

Explanation of Solution

Given:

The volume is $v=x{\left(3-2x\right)}^2$

Calculation:

$v=x{\left(3-2x\right)}^2$

Differentiate with respect to $x$

$\begin{array}{c}\frac{dV}{dx}=x\times 2\times \left(3-2x\right)\times \left(-2\right)+{\left(3-2x\right)}^2\times 1\\ =\left(3-2x\right)\times \left[-4x+3-2x\right]\\ =\left(3-2x\right)\times \left[3-6x\right]\end{array}$

For critical points, $\frac{dV}{dx}=0$

$\left(3-2x\right)\times \left[3-6x\right]=0$

Either $x=\frac{3}{2}$ or $x=\frac{1}{2}$

Differentiate $\frac{dV}{dx}$ with respect to $x$

$\begin{array}{c}\frac{d^{2}V}{d{x}^2}=\left(3-2x\right)\times \left(-6\right)+\left(3-6x\right)\times \left(-2\right)\\ =-18+12x-6+12x\\ =-24+24x\end{array}$

Substitute the value of $x=\frac{1}{2}$ in $\frac{d^{2}V}{d{x}^2}$

$\begin{array}{c}\frac{d^{2}V}{d{x}^2}=-24+24\times \frac{1}{2}\\ =-12<0\end{array}$

$x=\frac{1}{2}$ yields the maximum volume.

Substitute the value of $x=\frac{3}{2}$ in $\frac{d^{2}V}{d{x}^2}$

$\begin{array}{c}\frac{d^{2}V}{d{x}^2}=-24+24\times \frac{3}{2}\\ =12>0\end{array}$

$x=\frac{3}{2}$ yields the minimum volume.

Substitute $x=\frac{1}{2}$ in $v=x{\left(3-2x\right)}^2$,

$\begin{array}{c}v=\frac{1}{2}{\left(3-2\frac{1}{2}\right)}^2\\ =\frac{1}{2}\times 2^2\\ =2\end{array}$

The calculated volume is equal to the estimated volume ${\text{2ft}}^3$.

Thus, the maximum volume is ${\text{2ft}}^3$.