Chapter 11.1, Problem 1MP

(a)

To determine

What is an algebraic expression for the volume of the box? Write an equation that relates this expression to the given volume 355 ${\text{in}}^{\text{3}}$ .

Answer to Problem 1MP

  $\text{Volume}\text{=}{\text{x}}^2\text{w}{\text{in}}^{\text{3}}$ and the required equation: ${\text{x}}^2\text{w}\text{=}\text{355}$

Explanation of Solution

Given:

What is an algebraic expression for the volume of the box? Write an equation that relates this expression to the given volume 355 ${\text{in}}^{\text{3}}$ .

The figure: $\text{Length}\text{=}\text{x}\text{in;}\text{Width}\text{=}\text{w}\text{in;}\text{Height}\text{=}\text{x}\text{in}$

  

Concept Used:

The volume of rectangular box is: $\text{Volume}\text{=}\text{Length}·\text{width}·\text{height}$

Calculation:

Given volume =355 ${\text{in}}^{\text{3}}$

  $\text{Volume}\text{=}\text{x}·\text{w}·\text{x}\text{=}{\text{x}}^2\text{w}{\text{in}}^{\text{3}}$

Required equation: ${\text{x}}^2\text{w}\text{=}\text{355}$

Thus, $\text{Volume}\text{=}{\text{x}}^2\text{w}{\text{in}}^{\text{3}}$ and the required equation: ${\text{x}}^2\text{w}\text{=}\text{355}$

(b)

To determine

Solve for w in the equation from part (a).

Answer to Problem 1MP

  $\text{w}=\frac{\text{355}}{{\text{x}}^2}$

Explanation of Solution

Given:

Solve for w in the equation from part (a).

The figure: $\text{Length}\text{=}\text{x}\text{in;}\text{Width}\text{=}\text{w}\text{in;}\text{Height}\text{=}\text{x}\text{in}$

  

Concept Used:

Required equation: ${\text{x}}^2\text{w}\text{=}\text{355}$

Calculation:

Required equation: ${\text{x}}^2\text{w}\text{=}\text{355}$

  $\begin{array}{l}{\text{x}}^2\text{w}\text{=}\text{355}\\ \frac{{\text{x}}^2\text{w}}{{\text{x}}^2}\text{=}\frac{\text{355}}{{\text{x}}^2}[\text{Divide}\text{each}\text{side}\text{by}{\text{x}}^2]\\ \text{w}=\frac{\text{355}}{{\text{x}}^2}\end{array}$

Thus, $\text{w}=\frac{\text{355}}{{\text{x}}^2}$

(c)

To determine

Write an expression in terms of x and w to represent the surface area of the lunch box.

Answer to Problem 1MP

  $\text{Surface}\text{area}=4\text{x}·\text{w}+2{\text{x}}^2$

Explanation of Solution

Given:

Write an expression in terms of x and w to represent the surface area of the lunch box.

The figure: $\text{Length}\text{=}\text{x}\text{in;}\text{Width}\text{=}\text{w}\text{in;}\text{Height}\text{=}\text{x}\text{in}$

  

Concept Used:

  $\text{Surface}\text{area}=2[\text{length}·\text{width}\text{+}\text{width}·\text{height}\text{+}\text{length}·\text{height}]$

  $\text{Length}\text{=}\text{x}\text{in;}\text{Width}\text{=}\text{w}\text{in;}\text{Height}\text{=}\text{x}\text{in}$

Calculation:

  $\text{Surface}\text{area}=2[\text{length}·\text{width}\text{+}\text{width}·\text{height}\text{+}\text{length}·\text{height}]$

  $\text{Length}\text{=}\text{x}\text{in;}\text{Width}\text{=}\text{w}\text{in;}\text{Height}\text{=}\text{x}\text{in}$

  $\text{Surface}\text{area}=2[\text{x}·\text{w}\text{+}\text{w}·\text{x}\text{+}{\text{x}}^2]$

  $\text{Surface}\text{area}=4\text{x}·\text{w}+2{\text{x}}^2$

Thus, $\text{Surface}\text{area}=4\text{x}·\text{w}+2{\text{x}}^2$

(d)

To determine

Substitute for w in the expression for the surface area of the lunch box, using your result from part (b). Simplify the expression.

Answer to Problem 1MP

  $\text{Surface}\text{area}=\frac{2}{x}({\text{x}}^3+710)$

Explanation of Solution

Given:

Substitute for w in the expression for the surface area of the lunch box, using your result from part (b). Simplify the expression. The figure: $\text{Length}\text{=}\text{x}\text{in;}\text{Width}\text{=}\text{w}\text{in;}\text{Height}\text{=}\text{x}\text{in}$

  

Concept Used:

  $\text{Surface}\text{area}=4\text{x}·\text{w}+2{\text{x}}^2$ and $\text{w}=\frac{\text{355}}{{\text{x}}^2}$

Calculation:

  $\text{Surface}\text{area}=4\text{x}·\text{w}+2{\text{x}}^2$ and $\text{w}=\frac{\text{355}}{{\text{x}}^2}$

  $\begin{array}{l}\text{Surface}\text{area}=4\text{x}·\frac{\text{355}}{{\text{x}}^2}\text{+}2{\text{x}}^2]=2{\text{x}}^2+\frac{1420}{\text{x}}=\frac{2}{x}({\text{x}}^3+710)\\ \text{Surface}\text{area}=\frac{2}{x}({\text{x}}^3+710)\end{array}$

Thus, $\text{Surface}\text{area}=\frac{2}{x}({\text{x}}^3+710)$