Problem 79 - Algebra and Trigonometry

Question

The perimeter of a rectangle is 50 feet. Describe the possible lengths of a side if the area of the rectangle is not to exceed 114 square feet.

Step-by-Step Solution

Answer

The possible lengths of the sides of the rectangle (to not exceed an area of 114 square feet) lie within [9, 14] feet for length and [11, 16] feet for the width.

Step 1: Understand the perimeter

The perimeter of a rectangle is given by 2(length + width). So, from 2(length + width) = 50, we can simplify this equation to length + width = 25. Suppose, let length = x, so width = 25 - x.

Step 2: Set up the inequality for the area

The area of the rectangle must not exceed 114 square feet, i.e., length*width ≤ 114. Substituting for width in the area inequality gives the inequation x * (25 - x) ≤ 114.

Step 3: Solve the inequality

Rearrange the inequality into a quadratic, resulting in \(x^2 - 25x + 114 \leq 0\). This quadratic factors into \((x - 9)(x - 14) \leq 0\). According to the properties of quadratic functions, the function is less than or equal to zero between its roots. So, the possible range of x (length) lies in the interval [9, 14]. The width, therefore, would be in the interval [25 - 14, 25 - 9], which corresponds to the interval [11, 16].

Step 4: Conclude

The possible lengths of the sides of the rectangle, so as not to exceed an area of 114 square feet, should be within [9, 14] feet for length and [11, 16] feet for width.