Problem 145 - College Algebra

Question

The length of a rectangular sign is 3 feet longer than the width. If the sign's area is 54 square feet, find its length and width.

Step-by-Step Solution

Answer

The width of the rectangular sign is 6 feet and the length is 9 feet.

Step 1: Define the variables

Let's identify what we know and define variables to represent unknowns. We know the area of the rectangle is 54 square feet and that the length is 3 feet longer than the width. Let's call the width \(w\) in feet and the length \(l\) in feet. Given the relationship of the length being 3 feet longer, \(l = w + 3\).

Step 2: Set up the equation

Now let's use the formula for the area of a rectangle, which is \(Area = Width x Length\) or \(A = w \cdot l\). Substituting the known values, we get \(54 = w \cdot (w + 3)\).

Step 3: Solve the equation

To solve the equation, we need to distribute \(w\) into \(w + 3\): \(54 = w^2 + 3w\). This is a quadratic equation. Rearrange to have 0 on one side: \(0 = w^2 + 3w - 54\).

Step 4: Factor the quadratic equation

We can factor the quadratic equation: \(0 = (w - 6)(w + 9)\).

Step 5: Solve for \(w\)

Setting each factor equal to zero gives the potential solutions \(w = 6\) and \(w = -9\). Since width can't be negative, we dismiss \(w = -9\) and so \(w = 6\) feet.

Step 6: Solve for \(l\)

Substitute \(w\) back into the linear equation from step 1: \(l = w + 3 = 6 + 3 = 9\) feet. Therefore, the length is 9 feet.