A College Is Creating A New Rectangular Parking Lot

A College Is Creating A New Rectangular Parking Lot The Length Is

A college is creating a new rectangular parking lot. The length is 0.08 mile longer than the width and the area of the parking lot is 0.024 square mile. Find the length and width of the parking lot. Show your work.

Paper For Above instruction

The problem involves determining the dimensions of a rectangular parking lot based on given properties. Let the width of the parking lot be denoted by \( w \). Since the length is 0.08 mile longer than the width, we can express it as \( l = w + 0.08 \). The area \( A \) of the rectangle is given as 0.024 square mile, which introduces the equation \( A = l \times w \). Substituting the expression for \( l \), we have \( 0.024 = (w + 0.08) \times w \). Expanding this, \( 0.024 = w^2 + 0.08w \). Rearranged to standard quadratic form, this becomes \( w^2 + 0.08w - 0.024 = 0 \).

Using the quadratic formula \( w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), with \( a = 1 \), \( b = 0.08 \), and \( c = -0.024 \), we calculate the discriminant:

\[

\Delta = (0.08)^2 - 4 \times 1 \times (-0.024) = 0.0064 + 0.096 = 0.1024

\]

Taking the square root:

\[

\sqrt{\Delta} = \sqrt{0.1024} = 0.32

\]

Now, solving for \( w \):

\[

w = \frac{-0.08 \pm 0.32}{2}

\]

This yields two solutions:

\[

w = \frac{-0.08 + 0.32}{2} = \frac{0.24}{2} = 0.12

\]

and

\[

w = \frac{-0.08 - 0.32}{2} = \frac{-0.40}{2} = -0.20

\]

Since a width cannot be negative, we discard \( w = -0.20 \) and accept \( w = 0.12 \). The length \( l \) is then:

\[

l = w + 0.08 = 0.12 + 0.08 = 0.20

\]

Therefore, the width of the parking lot is 0.12 miles, and the length is 0.20 miles. These dimensions satisfy the area requirement, as:

\[

0.12 \times 0.20 = 0.024 \text{ square miles}

\]

The calculated dimensions correctly reflect the problem's conditions and provide the solution clearly.

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