Question

A running track of 440 ft is to be laid out enclosing a football field, the shape of which is a rectangle with a semi-circle at each end. If the area of the rectangular portion is to be maximum, then the lengths of its sides are:

• A. 70 ft and 110 ft
• B. 80 ft and 120 ft
• C. 35 ft and 110 ft
• D. 35 ft and 120 ft

Hint:

When maximizing the area of a geometric shape, use the perimeter constraint to express one dimension in terms of the other and then apply calculus to find the optimal dimensions.

Answer 1

The Correct Option is C

Maximizing the Area of a Combined Rectangular and Circular Shape 
Given that the perimeter is 440 ft, we define the dimensions as follows: Let \( x \) be the length of the rectangle, Let \( r \) be the radius of the semicircle. Since the perimeter includes two sides of the rectangle and the circumference of the semicircle, we have: \[ 2x + 2\pi r = 440. \] Simplifying for \( r \), we find: \[ 2x + 2\pi r = 440 \implies r = \frac{220 - x}{\pi}. \] The area \( A \) of the shape, consisting of the area of the rectangular part and the semicircular part, is: \[ A = xr + \frac{1}{2}\pi r^2. \] Substituting \( r \) from the perimeter equation: \[ A = x\left(\frac{220 - x}{\pi}\right) + \frac{1}{2}\pi \left(\frac{220 - x}{\pi}\right)^2. \] To find the maximum area, we set the derivative of \( A \) with respect to \( x \) to zero: \[ \frac{dA}{dx} = \frac{1}{\pi}(440x - 2x^2 - 440x + 2x^2) = 0, \] which simplifies and yields: \[ x = 110 \quad {(checking the second derivative confirms a maximum)}. \] At \( x = 110 \), we compute \( r \) as: \[ r = \frac{220 - 110}{\pi} = \frac{110}{\pi}. \] Converting to a more practical form, we find: \[ r = \frac{440 - 220}{\frac{22}{7}} \approx 35 { ft}, \] and thus: \[ x = 110 { ft}. \] Conclusion: The maximum area configuration occurs when \( x = 110 { ft} \) and \( r \approx 35 { ft} \).