Problem 57 - College Algebra

Question

Find the length and width of a rectangle whose perimeter is 36 feet and whose area is 77 square feet.

Step-by-Step Solution

Answer

The dimensions of the rectangle are width 7 feet and length 11 feet or width 11 feet and length 7 feet, depending on which side is defined as width or length.

Step 1: Write down the given equations

The perimeter is represented by \(2 \times (length + width) = 36\), and the area by \(length \times width = 77\). Therefore, we have two equations as follows: \((length + width) = 18\) and \(length \times width = 77\).

Step 2: Solve the first equation for one variable

Let's solve \((length + width) = 18\) for width: \(width = 18 - length\).

Step 3: Substitute the value of width in the second equation

Substitute \(width = 18 - length\) into the area equation. This gives us \(length \times (18 - length) = 77\). This simplifies to \(-length^2 + 18length = 77\). Shifting all terms to one side to form a quadratic equation yields \(length^2 - 18length + 77 = 0\).

Step 4: Solve the quadratic equation

The quadratic formula is \(length = \frac{-b \pm \sqrt{b^2-4ac}}{2a}\). Substitute a = 1, b = -18, and c = 77 into the quadratic formula to get the lengths. Remember that the lengths are positive since they represent physical dimensions. So, we discard the negative root. Solving gives us \(length = 7\) and \(length = 11\).

Step 5: Find the width corresponding to each length

Substitute each of the lengths back into the equation \(width = 18 - length\) to find corresponding widths. For \(length = 7\), we have \(width = 11\), and for \(length = 11\), we have \(width = 7\). So, the dimensions of the rectangle can either be 7ft by 11ft or 11ft by 7ft, depending on which side we coin as width or length.