The dimensions of a playground are 50 feet { \times } 75 feet. You want to double its area by...

Question:

The dimensions of a playground are 50 feet {eq}\times {/eq} 75 feet. You want to double its area by adding the same distance x to the length and width. Find x and the new dimensions of the playground.

Area of a Rectangle

Rectangle is a shape with four right angles having two pairs of opposite equal sides. The area of a rectangle is found by multiplying its length by its width. That said, having a length of a and a width of b, the area of such a rectangle would be defined as {eq}A = ab {/eq}.

The playground might be modeled as a rectangle of length a = 75 ft and width b = 50 ft. Initially, we could find the area of the playground using the area formula for the rectangle:

{eq}A_1 = ab = 75 ft\cdot 50 ft = 3750 ft^2 {/eq}

The problem states that we wish to double the area. We may then multiply the initial area by 2 to obtain the value for the desired area:

{eq}A_2 = 2A_1 = 2\cdot 3750 ft^2 = 7500 ft^2 {/eq}

This new rectangular playground would have both its length and width increased by x units. Let's define the new length and width based on that:

{eq}L = 75 + x\\ W = 50 + x {/eq}

Multiplying the new length and the new width would produce the desired area we've calculated above:

{eq}L\cdot W = (75 + x)(50 + x) = 7500 {/eq}

Now all we need is to solve for x. To do this, let's foil and combine the like terms, then rearrange into a standard quadratic form:

{eq}3750 + 75x + 50x + x^2 = 7500\\ x^2 + 125x - 3750 = 0 {/eq}

In order to solve this equation, we will apply the quadratic formula. For an equation in a form {eq}ax^2 + bx + c = 0 {/eq}, the two solutions are given by:

{eq}x_{1, 2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} {/eq}

For our equation: a = 1, b = 125, c = -3750. Substituting these values into the formula, we get:

{eq}x_{1, 2} = \frac{-125 \pm \sqrt{125^2 - 4\cdot 1\cdot (-3750)}}{2\cdot 1} = \frac{-125 \pm 175}{2} {/eq}

We're only interested in the positive solution, as x > 0, the area increases:

{eq}x = \frac{-125 + 175}{2} = 25 {/eq}

The new dimensions are then:

L = 75 + x = 100 ft and W = 50 + x = 75 ft.