Chapter 9.5, Problem 40PPS

To determine

Find the solution.

Answer to Problem 40PPS

Length l = $80$ in

Width w = $6$ in

Explanation of Solution

Given Information: Area of Canvas = $480i{n}^2$ , Width w = $4+x$ in., Length l = $60+10x$ in.

Formula Used : Area of Rectangle = $Lenght\times Width$ , $a^2+2ab+b^2={(a+b)}^2$ and

  $a^2-b^2=(a+b)(a-b)$

Calculation:

Width w = $4$ in. and increase in Length l = $60$ in.

Let increase in Width w = $x$ in. and increase in Length l = $10x$ in.

Area of Canvas = $480i{n}^2$ , Width w = $4+x$ in., Length l = $60+10x$ in

Area of Rectangle = $Lenght\times Width$

  $480=\left(60+10x\right)\times \left(4+x\right)$

So,

  $480=240+60x+40x+10x^2$

Now, simplify

  $10x^2+100x-240=0$

Now, to make coefficient of $x^2$ = 1, divide the equation by 10.

  $x^2+10x-24=0$

Now, adding & subtracting the square of half the coefficient of x.

  $x^2+10x-24+{\left(\frac{10}{2}\right)}^2-{\left(\frac{10}{2}\right)}^2=0$

Now use formula, $a^2+2ab+b^2={(a+b)}^2$

  ${\left(x+5\right)}^2-24-{\left(5\right)}^2=0$

So,

  ${\left(x+5\right)}^2-24-25=0$

So,

  ${\left(x+5\right)}^2-49=0$

Now, square root of 49 is 7

  ${\left(x+5\right)}^2-7^2=0$

Now use formula, $a^2-b^2=(a+b)(a-b)$

  $\left(x+5+7\right)\left(x+5-7\right)=0$

Now,

  $\left(x+12\right)\left(x-2\right)=0$

Hence,

$x=-12$ or $x=2$

Length l = $60+10x=60+10\times 2=60+20=80$ in

Width w = $4+x=4+2=6$ in