Chapter 12.1, Problem 3E

(a)

To determine

To express: The volume of the box in terms of $x$ .

Explanation of Solution

Given information: Square corners cuts out of pieces of metal that measure $8\text{in}$ by $15\text{in}$ .

Calculation:

The diagram of a metal sheet that measure $8\text{in}$ by $15\text{in}$ is shown below. The Square corner of side $x\text{in}$ cuts out from each corner.

  

Figure (1)

So, the box from metal sheet can be made as:

  

Figure (2)

The box is in the shape of a cuboid. So, use the formula to calculate volume of cuboid is:

  $\text{Volume}=\text{length}\cdot \text{width}\cdot \text{height}$

To find the volume in terms of x , substitute $15-2x$ for length, $8-2x$ for width and $x$ for height in the above formula.

  $\text{Volume}=\left(15-2x\right)\cdot \left(8-2x\right)\cdot x$

Therefore the volume of the box in terms of x is $x\left(15-2x\right)\left(8-2x\right)$ .

(b)

To determine

To find: The maximum volume of the box to the nearest tenth of cubic inch.

Answer to Problem 3E

The maximum volume of the box to the nearest tenth is $90.7{\text{in}}^{\text{3}}$ .

Explanation of Solution

Given information: Square corners cuts out of pieces of metal that measure $8\text{in}$ by $15\text{in}$ .

Calculation:

From part (a), the volume of the box is:

  $\begin{array}{c}\text{Volume}=\left(15-2x\right)\cdot \left(8-2x\right)\cdot x\\ V=4x^3-46x^2+120x\end{array}$

Now, differentiate the function of volume.

  $V^{\prime }\left(x\right)=12x^2-92x+120$

Equate the function to zero.

  $\begin{array}{c}{V}^{\prime }\left(x\right)=0\\ 12x^2-92x+120=0\\ x=6,\frac{5}{3}\end{array}$

Again differentiate the function to find the maximum.

  $V^{″}\left(x\right)=24x-92$

The function given negative value for $x=\frac{5}{3}$ . So, the maximum volume of the box can be calculated as:

  $\begin{array}{c}V=\left(15-\frac{10}{3}\right)\cdot \left(8-\frac{10}{3}\right)\cdot \frac{5}{3}\\ =90.7407{\text{in}}^3\end{array}$

Therefore, the maximum volume of the box to the nearest tenth is $90.7{\text{in}}^{\text{3}}$ .

(c)

To determine

To find: The length, width, and height of the box to nearest tenth.

Answer to Problem 3E

The length of the box is 11.6 in, breath is 4.6 in and height is 1.7 in.

Explanation of Solution

Given information: Square corners cuts out of pieces of metal that measure $8\text{in}$ by $15\text{in}$ .

Calculation:

From part (b), the maximum volume of the box is $90.7{\text{in}}^{\text{3}}$ . The function for the volume is $V\left(x\right)=\left(15-2x\right)\cdot \left(8-2x\right)\cdot x$ . So,

  $\begin{array}{c}V\left(x\right)=90.7\\ \left(15-2x\right)\cdot \left(8-2x\right)\cdot x=90.7\\ x=1.7\text{in}\end{array}$

So, the height of the box is 1.7 in.

The length of the box is:

  $\begin{array}{c}\text{Length}=15-2x\\ =15-2\left(1.7\right)\\ =15-3.4\\ =11.6\text{in}\end{array}$

The breath of the box is:

  $\begin{array}{c}\text{Breadth}=8-2x\\ =8-2\left(1.7\right)\\ =4.6\text{in}\end{array}$

Therefore, the length of the box is 11.6 in, breath is 4.6 in and height is 1.7 in.