Q3. - Geometry

Question

Suppose the manufacturing company cuts square corners out of pieces of metal that measure 8 inches by 15 inches.

a.Express the volume in terms of x.

b.Find the maximum volume, correct to the nearest tenth of a cubic inch.

c.What are the length, width and height of the box with maximum volume? Give each correct to the nearest tenth of an inch.

Step-by-Step Solution

Answer

1. The volume is$x(15-2x)(8-2x)$ .
2. The maximum volume is$90.7{\text{\hspace{0.17em}in}}^{\text{3}}$ .
3. The length of the box is 11.6 in, breath is 4.6 in and height is 1.7 in.

Step 1. Given information.

Square corners cut out of pieces of metal that measure 8 inches by 15 inches.

Step 2. Write the concept.

The diagram of a metal sheet that measure 8in by 15inis shown below. The Square corner of side x inch cuts out from each corner.

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

Step 3. Determine the volume.

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.

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

 

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

Step 1. Given information.

Square corners cut out of pieces of metal that measure 8 inches by 15 inches.

Step 2. Explanation.

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

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

 

Now, differentiate the function of volume.

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

 

Equate the function to zero.

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

Step 3. Determine the maximum volume.

Again, differentiate the function to find the maximum.

$V^{\text{'}\text{'}}\left(x\right)=24x-92$

 

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

$\begin{array}{c}V=(15-\frac{10}{3})\cdot (8-\frac{10}{3})\cdot \frac{5}{3}\\ =90.7407{\text{\hspace{0.17em}in}}^3\end{array}$

 

Therefore, the maximum volume of the box to the nearest tenth is $90.7{\text{\hspace{0.17em}in}}^{\text{3}}$.

Step 1. Given information.

Square corners cut out of pieces of metal that measure 8 inches by 15 inches.

Step 2. Explanation.

From part (b), the maximum volume of the box is$90.7{\text{\hspace{0.17em}in}}^{\text{3}}$ .

The function for the volume is $V\left(x\right)=(15-2x)\cdot (8-2x)\cdot x$. So,

 $\begin{array}{c}V\left(x\right)=90.7\\ (15-2x)\cdot (8-2x)\cdot x=90.7\\ x=1.7\text{\hspace{0.17em}in}\end{array}$

So, the height of the box is$1.7\text{\hspace{0.17em}in}$ .

Step 3. Determine values.

The length of the box is:

$\begin{array}{c}\text{Length}=15-2x\\ =15-2\left(1.7\right)\\ =15-3.4\\ =11.6\text{\hspace{0.17em}in}\end{array}$

The breath of the box is:

 $\begin{array}{c}\text{Breadth}=8-2x\\ =8-2\left(1.7\right)\\ =4.6\text{\hspace{0.17em}in}\end{array}$

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