Chapter 9.4, Problem 48PPS

To determine

To calculate: The size of Pool should Ichiro have built to fit his budget.

Answer to Problem 48PPS

The size of the Pool is 350 inches by 350 inches by 42 inches.

Explanation of Solution

Given Information:

The swimming pool of model A is 42 inches deep and holds cubic feet of water. The length of the pool is 5 feet more than the width. Ichiro has budgeted $200 per month to heat the pool. His neighbor owns model A and she spends about $150 per month to her pool. Ichiro wants a larger pool with a depth of 42 inches changing the length and breadth by a combined 10 feet increases the heating cost by 25%.

Ichiro has budgeted 200 per month to heat his poo and his neighbor pays 15 per month.

Ichiro budget

  $\begin{array}{l}=\frac{200-150}{150}\times 100\\ =33\frac{1}{3}\%\end{array}$

By increasing the length and breadth by a combined 10 feet increase the cost by 25% so increasing the length and breadth by 1 foot increases the cost by 25%. Because Ichiro wants the cost to increase by $33\frac{1}{3}\%$ .

The length and width of his pool can increase by a combined amount:

  $\frac{33\frac{1}{3}}{2.5}=\frac{40}{3}$

Now,

let width of pool = $w\text{ ft}$

Length of the pool = $\left(w+5\right)\text{ ft}$

Height of the pool $=42\text{ inches}=3.5\text{ ft}$

  $\begin{array}{c}\text{Volume}=1750\\ LBH=1750\\ (w+5)w(3.5)=1750\\ (w^2+5w)=\frac{1750}{3.5}\\ w^2+5w=\frac{17500}{35}=500\\ w^2+5w-500=0\\ w^2+25w-20w-500=0\\ w(w+25)-20(w+25)=0\\ (w-20)(w+25)=0\\ w=20,-25\end{array}$

-25 is not possible

Thus, the $w=20$

Length will be $L=20+5=25$

Length and width can increase combined by $\frac{40}{3}$

Let width increase = $x$

Length increase = $\frac{40}{3}-x$

New width = $20+x$

New length = $25+\frac{40}{3}-x$

  $=\frac{115}{3}-x$

Volume = LBH

  $\begin{array}{c}\text{Volume}=1750\\ LBH=1750\\ (w+5)w(3.5)=1750\\ (w^2+5w)=\frac{1750}{3.5}\\ w^2+5w=\frac{17500}{35}=500\\ w^2+5w-500=0\\ w^2+25w-20w-500=0\\ w(w+25)-20(w+25)=0\\ (w-20)(w+25)=0\\ w=20,-25\\ 25+\frac{40}{3}-x\\ =\frac{115}{3}-x\\ =(\frac{115}{.3}-x)(20+x)(3.5)\\ =3.5[\frac{2300}{3}+\frac{115}{3}x-20x-x^2]\\ =\frac{35}{10}[\frac{2300+115x-60x-3x^2}{3}]\\ =\frac{35}{10}[\frac{2300+55x-3x^2}{3}]\\ =\frac{1}{30}[80500+1925x-105x^2]\end{array}$

Calculate the x coordinate of the pool of vertex parabola.

Then

  $\begin{array}{c}\text{width}=20+x=20+\frac{55}{6}\\ =\frac{175}{6}\text{ft}\\ \text{length}=\frac{115}{3}-\frac{55}{6}\\ =\frac{230-55}{6}=\frac{175}{6}\text{ft}\\ =\frac{175}{6}\times 12\\ =350\text{in}\end{array}$

  $\text{Width }=\frac{175}{6}=350\text{in}$

The size of the pool is 350 inches by 350 inches by 42 inches