Chapter 8.1, Problem 50E

To determine

To find: the dimensions of original box.

Explanation of Solution

Given information:

Let $x$ represents the width of the original box.

Dimensions of original box are as follows:

  $\begin{array}{l}\text{Width}=x\\ \text{Height}=2x\\ \text{Length}=x+1\end{array}$

Dimensions of new box are as follows:

  $\begin{array}{l}\text{Width}=x+1\\ \text{Height}=2x+2\\ \text{Length}=x+2\end{array}$

Volume of new box $\left(V_{new}\right)=160\text{ cubic feet}$

Calculation:

First evaluate the value of $x$ as shown below.

  $\begin{array}{c}{V}_{new}=\left(\text{Width}\right)\left(\text{Height}\right)\left(\text{Length}\right)\\ 160=\left(x+1\right)\left(2x+2\right)\left(x+2\right)\left[\text{Put the dimesnsions of new box}\right]\\ 160=2x^3+8x^2+10x+4\\ 0=x^3+4x^2+5x-78\end{array}$

If the above equation has any ration solution it will have a factor of 78. The above equation has a 3 as a ration solution, i.e. $x=3$ .

Now divide the polynomial equation by $x-3$ to check whether it has more solutions or not.

  $\frac{x^3+4x^2+5x-78}{x-3}=x^2+7x+26$

Thus the above equation has only one solution i.e. $x=3$ .

The dimensions of original box can be evaluated by plugging $x$ into the equations shown below.

  $\begin{array}{c}\text{Width}=x\\ =3\text{ feer}\\ \text{Height}=2x\\ =2\left(3\text{ feet}\right)\\ =6\text{ feet}\\ \text{Length}=x+1\\ =3+1\\ =4\text{ feet}\end{array}$

Thus the width, height and length of original box is 3 feet, 6 feet and 4 feet respectively.