Chapter 12.5, Problem 23STP

To determine

To find: which cardboard box does not meet the requirement.

Answer to Problem 23STP

The correct choice is (D).

Explanation of Solution

Given:

The dimension of box whose surface area is more is 6 in. by 14 in. by 20 in.

A) 7 in. by 10 in. by 24 in

B) 7 in. by 12 in. by 20 in

C) 10 in. by 12 in. by 14 in

D) 5 in. by 16 in. by 21 in

Calculation:

Consider two cardboard has same volume but one has less surface area than the other.

The dimension of box whose surface area is more is 6 in. by 14 in. by 20 in.

The objective is to find the size of box from the given box which does not meet the above criteria.

Consider l represents the length, w represents the width and h represents the height of a rectangular prism.

Find the surface area of the rectangular prism.

The lateral area L of a rectangular prism is, L = Ph.

Here, P is the perimeter of the base and h is the height the rectangular prism.

The surface area S of a rectangular prism is, S = L + 2B .

Here, L is the lateral area and B is the area of the base.

The perimeter P of the base of the prism is, $P=2\left(l+w\right)$ .

Here, l is the length and w is the width of the rectangular base.

Thus,

  $\begin{array}{c}L=P\cdot h\\ =2\left(l+w\right)\end{array}$

Find the surface area of the original prism as,

First find the area B of the base of the prism.

The area B of the base is, $B=l\cdot w$ .

Here, l is the length and w is the width of the rectangular base.

Thus, the surface area S of the original rectangular prism is

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

Therefore, the surface area of the rectangular prism is $S=2\left(lh+wh+lw\right)$ .

The volume V of the rectangular box with dimension 6 in. by 14 in. by 20 in. is,

  $\begin{array}{c}V=l\times w\times h\\ =6\times 14\times 20\\ =1680{\text{in}}^3\end{array}$

The surface area S of the rectangular box with dimension 6 in. by 14 in. by 20 in. is,

  $\begin{array}{c}S=2\left(lh+wh+lw\right)\\ =2\left(6\cdot 20+14\cdot 20+6\cdot 14\right)\\ =2\left(120+280+84\right)\\ =2\left(484\right)\\ =968\end{array}$

The volume $V_1$ of the rectangular box with dimension 7 in. by 10 in. by 24 in. is,

  $\begin{array}{c}{V}_1=l\times w\times h\\ =7\times 10\times 24\\ =1680{\text{in}}^3\end{array}$

The surface area $S_1$ of the rectangular box with dimension 7 in. by 10 in. by 24 in. is,

  $\begin{array}{c}{S}_1=2\left(lh+wh+lw\right)\\ =2\left(7\cdot 24+10\cdot 24+7\cdot 10\right)\\ =2\left(168+240+70\right)\\ =2\left(478\right)\\ =956\end{array}$

Thus, $S_1<S$ , hence the dimension 7 in. by 10 in. by 24 in. meet the criteria.

The volume $V_2$ of the rectangular box with dimension 7 in. by 12 in. by 20 in. is,

  $\begin{array}{c}{V}_2=l\times w\times h\\ =7\times 12\times 20\\ =1680{\text{in}}^3\end{array}$

The surface area of the rectangular box with dimension 7 in. by 12 in. by 20 in. is,

  $\begin{array}{c}{S}_2=2\left(lh+wh+lw\right)\\ =2\left(7\cdot 20+12\cdot 20+7\cdot 12\right)\\ =2\left(140+240+84\right)\\ =2\left(464\right)\\ =928\end{array}$

Thus, $S_2<S$ , hence the dimension 7 in. by 12 in. by 20 in. meet the criteria.

The volume $V_3$ of the rectangular box with dimension 10 in. by 12 in. by 14 in. is,

  $\begin{array}{c}{V}_3=l\times w\times h\\ =10\times 12\times 14\\ =1680{\text{in}}^3\end{array}$

The surface area $S_3$ of the rectangular box with dimension 10 in. by 12 in. by 14 in. is,

  $\begin{array}{c}{S}_3=2\left(lh+wh+lw\right)\\ =2\left(10\cdot 14+12\cdot 14+10\cdot 12\right)\\ =2\left(140+168+120\right)\\ =2\left(428\right)\\ =856\end{array}$

Thus, $S_3<S$ , hence the dimension 10 in, by 12 in. by 14 in. meet the criteria.

The volume $V_4$ of the rectangular box with dimension 5 in. by 16 in. by 21 in. is,

  $\begin{array}{c}{V}_4=l\times w\times h\\ =5\times 16\times 21\\ =1680{\text{in}}^3\end{array}$

The surface area $S_4$ of the rectangular box with dimension 5 in. by 16 in. by 21 in. is,

  $\begin{array}{c}{S}_4=2\left(lh+wh+lw\right)\\ =2\left(5\cdot 21+16\cdot 21+5\cdot 16\right)\\ =2\left(105+336+80\right)\\ =2\left(521\right)\\ =1042\end{array}$

Thus, $S_4<S$ , hence the dimension 5 in. by 16 in. by 21 in. do not meet the criteria.

Conclusion:

Therefore, the correct choice is (D).