Problem 49 - 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 length and width of the rectangle could be 7 feet and 11 feet or 11 feet and 7 feet.

Step 1: Write down the formulas for the perimeter and the area of rectangle

Perimeter = 2L + 2W, Area = L*W. These are the formulas for the rectangle's perimeter and area, where L is the length and W is the width.

Step 2: Substitute the given perimeter and area into the formulas

We know that the perimeter is 36, so we can set up equation 2L + 2W = 36. And the area is 77, so we get L*W = 77.

Step 3: Solve the two equations

First, rewrite the perimeter equation into 2L = 36 - 2W or L = 18 - W. This is used because it can be easily substituted into the area equation. Substitute L into the area equation, we have: (18 - W)*W = 77. Opening brackets to simplify, we get: 18W - W^2 = 77. Then rearrange to find the quadratic equation: W^2 - 18W + 77 = 0.

Step 4: Solve the quadratic equation for W

Solving this quadratic equation will provide two solutions for the width (W), which are 11 and 7. To check which one is correct, substitute each found width into the area equation (77 = L*W) and verify if any of them give a valid length (L).

Step 5: Verify the width and length against the given parameters

Plugging both 11 and 7 as width into the area equation (77=L*W), length gets values 7 and 11 respectively. This means, two possible options for length and width are (11,7) and (7,11). Both options satisfy the condition for the perimeter 2L+2W=36. Therefore, there are two sets of potential values for L and W which are (7 feet, 11 feet) and (11 feet, 7 feet).