Chapter 12.5, Problem 17PPS

To determine

The higher material cost to make the box from following two box options:

  

Answer to Problem 17PPS

The higher material cost to make requires by the box 1.

Explanation of Solution

Given:

  

  $\begin{array}{l}{w}_1=2.5in\\ l_1=3.8in\\ h_1=1.4in\\ w_2=2.4in\\ l_2=2.4in\\ h_2=2.4in\\ M=\$1.30/i{n}^2\end{array}$

Concept Used:

The surface area of the rectangular prism is given by the following expression:

  $SA=2\left(wl+hl+hw\right)$

Where,

Length of rectangular prism: $l$

Widthof rectangular prism: $w$

Height of rectangular prism: $h$

Calculation:

The surface area of the rectangular prismor box 1:

  $\begin{array}{l}S{A}_1=2\left(w_1{l}_1+h_1{l}_1+h_1{w}_1\right)\\ S{A}_1=2\left[\left(2.5\times 3.8\right)+\left(1.4\times 3.8\right)+\left(1.4\times 2.5\right)\right]\\ S{A}_1=36.64i{n}^2\end{array}$

The surface area of the rectangular prism or box 2:

  $\begin{array}{l}S{A}_2=2\left(w_2{l}_2+h_2{l}_2+h_2{w}_2\right)\\ S{A}_2=2\left[\left(2.4\times 2.4\right)+\left(2.4\times 2.4\right)+\left(2.4\times 2.4\right)\right]\\ S{A}_2=28.8i{n}^2\end{array}$

Know that:

The material cost $\left(M\right)$ for $1i{n}^2$ is $\$1.30$ .

For box 1: $M_1=36.64\times 1.30=\$47.63/i{n}^2$

For box 2: $M_2=28.8\times 1.30=\$37.44/i{n}^2$

It clearly shows that the box 1 will contain the high material cost to make.

Conclusion:

Hence, the surface area of the box 1 is more that’s why it will require more material cost.