Chapter 9.5, Problem 50E

To determine

To find: the dimensions of the original pool.

Answer to Problem 50E

The dimensions of the original pool are $5,25,50$ .

Explanation of Solution

Given information:

Thevolume of the new pool will be $3240\text{ft}{\text{.}}^3$ and original pool has a width five times its depth and length ten times its depth. The new pool has increasing width of the original pool by four feet and length by six feet and depth by two feet.

Calculation: let depth of original pool is $x$ , then the width of the original pool is $5x$ and length is $10x$ ,

Since, The new pool has increasing width of the original pool by four feet and length by six feet and depth by two feet, so the dimensions of the new pool will be $10x+6,5x+4,x+2$ .

So the volume of new pool,

  $\begin{array}{l}\left(10x+6\right)\left(5x+4\right)\left(x+2\right)=3420\\ 50x^2+70x+24\left(x+2\right)=3420\text{ [multiply]}\\ 50x^3+70x^2+24x+100x^2+140x+48=3420\text{ [multiply]}\\ 50x^3+170x^2+164x-3372=0\text{ [add and subtract]}\\ 25x^3+85x^2+82x-1686=0\text{ [divide by 2]}\\ x=3\text{ [simplify]}\end{array}$

Thus, the dimensions of the original pool are $5,25,50$ .