Chapter 1.1, Problem 59E

To determine

To express: The volume V of the box as a function of x.

Answer to Problem 59E

The volume of the box as a function of x is $\underset{\_}{V\left(x\right)=4x^3-64x^2+240x}$

Explanation of Solution

Given:

Dimension of the rectangular piece of cardboard is 12 in. by 20 in.

Side length of the square is x in.

Formula used:

Volume of the rectangle of length l, breadth b and height h is, $V=l\times b\times h$.

Area of the rectangle of length l and breadth b is $A=l\times b$.

Calculation:

Since a square of side length x in. is cut out from each corner, the length and the width of the box are $l=20-x-x,w=12-x-x$.

Hence, the volume of the rectangular box is calculated as follows.

$\begin{array}{c}V\left(x\right)=\left(20-2x\right)\times \left(12-2x\right)\times x\\ =4x\left(10-x\right)\times \left(6-x\right)\\ =4x\left(60-16x+x^2\right)\\ =4x^3-64x^2+240x\end{array}$

Therefore, the volume of the box as a function of x is $V\left(x\right)=4x^3-64x^2+240x$.

As the dimension of the box cannot take negative values, $l>0,b>0,x>0$.

Case (i): $l>0$

$\begin{array}{l}20-2x>0\\ 2\left(10-x\right)>0\\ 10-x>0\\ x<10\end{array}$

Case (ii): $b>0$

$\begin{array}{l}12-2x>0\\ 2\left(6-x\right)>0\\ 6-x>0\\ x<6\end{array}$

Thus, the domain of the volume function $V\left(x\right)$ is $0<x<6$.