Chapter 6.5, Problem 71PPS

a.

To determine

Tofind:thevalue of x if the total area of the pool is 336 square feet.

Answer to Problem 71PPS

  $t=0,-35,15$ .

Explanation of Solution

Given:

The pool is surrounded by a sidewalk of the constant width is shown in given figure.

  

Calculation:

According to the figure,the dimension of the pool is:

  $\begin{array}{l}AB=12-2x\\ AF=28-2x\\ EG=CF=16-2x\\ DE=CE-CD=24-2x-\left(12-2x\right)=12\\ AC=AF-CF=28-2x-\left(16-2x\right)=12\end{array}$

AS,

  $\begin{array}{l}12-2x\ge 0\Rightarrow x\le 6\\ 28-2x\ge 0\Rightarrow x\le 14\\ 24-2x\ge 0\Rightarrow x\le 12\\ \therefore x\in \left[0,6\right]\end{array}$

Therefore,

  $\begin{array}{l}P\left(t\right)=-0.012t^3-0.24t^2+6.3t+8000\left\{\text{given}\right\}\\ 8000=-0.012t^3-0.24t^2+6.3t+8000\left\{\text{substitutes for }P\left(t\right)\right\}\\ 0.012t^3+0.24t^2-6.3t=0\{\text{subtract 8000 from both sides}}\\ 12t^3+240t^2-6300t=0\{\text{multiply 1000 from both sides}}\\ 12t\left(t^2+20t-525\right)=0\{\text{taking }12t\text{ as common factor}}\mathrm{.....}\text{{1}}\end{array}$

Now, to find the area of the pool is

  $AC\cdot CD+CF\cdot FG=336$

  $12\left(12-x\right)+\left(16-2x\right)\left(24-2x\right)=336\left\{\text{substitutes these value}\right\}$

  $144-12x+384-32x-48x+4x^2-336=0\{\text{simplify}}$

  $4x^2-104x+192=0\text{{add like terms}\}$

  $4\left(x^2-26x+48\right)=0\text{{take 4 as commom factor}}$

  $x^2-2x-24x+48=0\text{{expand}\}$

  $x\left(x-2\right)-24\left(x-2\right)=0\text{{take }\left(x-2\right)\text{ as commom factor}}$

  $\left(x-2\right)\left(x-24\right)=0\text{{simplify}\}$

Either $\left(x-2\right)=0$ or $\left(x-24\right)=0$

When $\left(x-2\right)=0\Rightarrow x=2$

When $\left(x-24\right)=0\Rightarrow x=24$

When $\left(t+15\right)=0\Rightarrow t=15$

Thus, the only value of $x=2$ because $x=24$ is not belong in that range.

Hence, the value of the polynomial equation is $x=2$ .

b.

To determine

To find:area of the new pool if the value of x is doubled.

Answer to Problem 71PPS

  $\text{176}{\text{ft}}^2$ .

Explanation of Solution

Given:

The area of the pool is 336 square feet and the value of x is doubled.

Calculation:

According to the question, the value of x is doubled, therefore the value of $x=4\left(2\times 2\right)$

  $\begin{array}{l}2AC\cdot CD+CF\cdot FG\\ =12\left(12-x\right)+\left(16-2x\right)\left(24-2x\right)\left\{\text{substitutes these value}\right\}\\ =12\left(12-4\right)+\left(16-2\times 4\right)\left(24-2\times 4\right)\{\text{simplify}}\\ =96+8\times 16\{\text{simplify}}\\ =96+128=176\end{array}$

Hence, the area of new pool is $\text{176}{\text{ft}}^2$

c.

To determine

To find: area of the new pool if the value of x is halved.

Answer to Problem 71PPS

  $\text{440}{\text{ft}}^2$ .

Explanation of Solution

Given:

The area of the pool is 336 square feet and the value of x is halved.

Calculation:

According to the question, the value of x is doubled, therefore the value of $x=1\left(2\times \frac{1}{2}\right)$

  $\begin{array}{l}2AC\cdot CD+CF\cdot FG\\ =12\left(12-x\right)+\left(16-2x\right)\left(24-2x\right)\left\{\text{substitutes these value}\right\}\\ =12\left(12-1\right)+\left(16-2\times 1\right)\left(24-2\times 1\right)\{\text{simplify}}\\ =12\times 11+14\times 22\{\text{simplify}}\\ =132+308=440\end{array}$

Hence, the area of new pool is $\text{440}{\text{ft}}^2$