Chapter 12.5, Problem 16PPS

(a)

To determine

To find: the surface area of the box.

Answer to Problem 16PPS

The surface area of the box is $432\text{in}{\text{.}}^2$

Explanation of Solution

Given:

Volume is 576 cubic inches

Length of the box is 12 in.

Width of the box is 8 in..

Formula used:

The surface area S of a rectangular pyramid is S = L + 2B

Calculation:

The objective is to find the surface area of the box.

The volume V of a rectangular box is given by

  $V=l\cdot w\cdot h$

Wherel is the length, w is the width and h is the height of the rectangular box.

The lateral area L of a rectangular pyramid is given by the formula

L = Ph

WhereP is the perimeter of the base and h is the height the rectangular pyramid.

The surface area S of a rectangular pyramid is given by the formula

S = L + 2B

WhereL is the lateral surface area and B is the area of the base.

On combining the last two formulas, the surface area is given by

  $\begin{array}{c}S=L+2B\\ =Ph+2B\\ =2\left(l+w\right)h+2lw\end{array}$

First find the height of the rectangular box.

Put V = 576, l = 12 and w = 8 in the volume formula to find height.

  $\begin{array}{c}V=l\cdot w\cdot h\\ 576=12\cdot 8\cdot h\\ =96h\\ 96h=576\\ h=\frac{576}{96}\left[\text{Divide}\text{both}\text{by}96\right]\\ =6\end{array}$

Now, find the surface area of the rectangular box.

Put l = 12, w = 8 and h = 6 in the surface area formula to find the surface area.

  $\begin{array}{c}S=2\left(l+w\right)h+2lw\\ =2\left(12+8\right)6+2\cdot 12\cdot 8\\ =2\left(20\right)6+192\\ =20\cdot 12+192\\ =240+192\\ =432\end{array}$

Conclusion:

Therefore, the surface area of the box is $432\text{in}{\text{.}}^2$ .

(b)

To determine

To state:which box has greater surface area.

Answer to Problem 16PPS

The second box has greater surface area than the first box.

Explanation of Solution

Given:

The dimensions of another box 16 in. by 9 in. by 4 in. whose volume is same as that of the first box.

Calculation:

The objective is to state which box has greater surface area.

Find the surface area of the second box.

Put l = 16, w = 9 and h = 4 in the combine formula for surface area to find the surface area of the second box.

  $\begin{array}{c}S=2\left(l+w\right)h+2lw\\ =2\left(16+9\right)4+2\cdot 16\cdot 9\\ =2\left(25\right)4+288\\ =25\cdot 8+288\\ =200+288\\ =488\end{array}$

Therefore, the surface area of the second box is $488\text{in}{\text{.}}^2$ .

Conclusion:

Therefore, the second box has greater surface area than the first box.

(c)

To determine

To check: whether the volume of the box is same as the last two boxes and the surface area is greater than the last two boxes

Answer to Problem 16PPS

The dimensions of the box 16 in. by 18 in. by 2 in. satisfy the condition.

Explanation of Solution

Calculation:

Consider the dimensions of another box 16 in. by 18 in. by 2 in.

The objective is to check whether the volume of the box is same as the last two boxes and the surface area is greater than the last two boxes.

The volume of the box is

  $\begin{array}{c}V=l\cdot w\cdot h\\ =16\cdot 18\cdot 2\\ =576\end{array}$

Therefore, the volume of the box is $576\text{in}{\text{.}}^3$ .

Put l = 16, w = 18 and h = 2 in the combine formula for surface area to find the surface area of the second box.

  $\begin{array}{c}S=2\left(l+w\right)h+2lw\\ =2\left(16+18\right)2+2\cdot 16\cdot 18\\ =2\left(34\right)2+576\\ =34\cdot 4+576\\ =136+576\\ =712\end{array}$

Therefore, the surface area of the box is $712\text{in}{\text{.}}^2$ .

Conclusion:

Therefore, the dimensions of the box 16 in. by 18 in. by 2 in. satisfy the condition.