Chapter 1, Problem 21REAE

Solution

To determine the value of $x$ that doubles the area of the rectangular playground when both the dimensions of the playground are added by $x$ feet.

It is found that:

  $x=24$ .

Given:

Given the dimensions of the rectangular play ground:

  $72\text{ft}\times \text{48}\text{ft}$ .

Given that $x$ feet are added to both the dimensions of the rectangular playground to double its area.

Concept Used:

Area of a rectangle:

Given the dimension of a rectangle: $a\text{unit}\times \text{ b}\text{unit}$ .

Then, the area of the rectangle:

  $ab{\text{unit}}^2$ .

Quadratic formula:

Given a quadratic equation $a{x}^2+bx+c=0$ .

Then, the solutions of the equation:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ .

Calculation:

Find the area of the playground $A_{old}$

before the expansion:

Find the area of the playground using the formula of the area of a rectangle:

  $\begin{array}{l}{A}_{old}=72\times 48{\text{ft}}^2\\ A_{old}=3456{\text{ft}}^2\end{array}$ .

That is, the area of the playground before the expansion:

  $A_{old}=3456{\text{ft}}^2$

Find the area of the playground $A_{new}$

after the expansion:

It is given that the area of the playground is doubled.

Thus, the area of the playground after the expansion:

  $\begin{array}{l}{A}_{new}=2A_{old}3456{\text{ft}}^2\\ A_{new}=2\cdot 3456{\text{ft}}^2\\ A_{new}=6912{\text{ft}}^2\end{array}$ .

Find the algebraic form of $A_{new}$in terms of $x$:

Since $x\text{ft}$ are added to both the dimensions of the rectangle, the new dimension of the rectangle:

  $\left(72+x\right)\text{ft}\times \left(48+x\right)\text{ft}$ .

Then, the area of the playground:

  $\begin{array}{l}{A}_{new}=\left(72+x\right)\left(48+x\right){\text{ft}}^2\\ A_{new}=x^2+120x+3456{\text{ft}}^2\end{array}$ .

Equate the algebraic form with the actual expanded area of the playground:

  $\begin{array}{l}6912{\text{ft}}^2=x^2+120x+3456{\text{ft}}^2\\ 6912=x^2+120x+3456\\ 0=x^2+120x-3456\end{array}$ .

Solve the equation $0=x^2+120x-3456$for $x$:

  $\begin{array}{l}x=\frac{-120\pm \sqrt{{120}^2-4\cdot 1\cdot \left(-3456\right)}}{2\cdot 1}\\ x=\frac{-120\pm \sqrt{28224}}{2}\\ x=\frac{-120\pm 168}{2}\\ x=\frac{-120+168}{2},\frac{-120-168}{2}\\ x=24,-144\end{array}$ .

Now, a negative value of $x$ is not feasible since $x$ represents an actual length in feet.

Thus, $x=24$ .

Conclusion:

It is found that:

  $x=24$ .