Question

If a running track of 500 ft. is to be laid out enclosing a playground, the shape of which is a rectangle with a semicircle at each end, then the length of the rectangular portion such that the area of the rectangular portion is maximum is (in feet).

• A. \( 100 \)
• B. \( 125 \)
• C. \( 150 \)
• D. \( 200 \)

Hint:

For maximum area problems with constraints, express the function in terms of a single variable and use differentiation to find the critical points.

Answer 1

The Correct Option is B

We are asked to maximize the area of the rectangular portion of a playground enclosed by a running track with a total perimeter of 500 ft. Step 1: Identify Variables Let the length of the rectangular portion be \( L \) and the width be \( 2R \), where \( R \) is the radius of the semicircles. Since there are two semicircles at each end, their total circumference equals the circumference of a full circle, which is \( 2\pi R \). Step 2: Perimeter Equation From the given total perimeter condition, \[ L + 2R + L + 2\pi R = 500 \] Simplifying, \[ 2L + (2 + 2\pi)R = 500 \] \[ 2L + 2(1 + \pi)R = 500 \] Dividing the entire equation by 2, \[ L + (1 + \pi)R = 250 \] Step 3: Area of the Rectangular Portion The area of the rectangular portion is: \[ A = L \times 2R \] From the perimeter equation: \[ L = 250 - (1 + \pi)R \] Now, \[ A = 2R[250 - (1 + \pi)R] \] Expanding: \[ A = 500R - 2(1 + \pi)R^2 \] Step 4: Maximizing the Area To maximize the area, take the derivative and set it equal to zero: \[ \frac{dA}{dR} = 500 - 4(1 + \pi)R \] Set the derivative equal to zero: \[ 500 - 4(1 + \pi)R = 0 \] \[ R = \frac{500}{4(1 + \pi)} \] Since \( \pi \approx 3.14 \), \[ R = \frac{500}{4 \times 4.14} = \frac{500}{16.56} \approx 30.2 \] Step 5: Finding \( L \) Using \( L = 250 - (1 + \pi)R \), \[ L = 250 - 4.14 \times 30.2 \] \[ L \approx 250 - 125 \] \[ L \approx 125 \] Step 6: Final Answer  Correct Answer: (2) \ 125 

Answer 2

The given problem involves a running track designed as a rectangle with a semicircle at each end. Our goal is to maximize the area of the rectangular portion while ensuring the total perimeter is 500 ft.

The rectangular playground has a width \( w \) and a length \( l \). The semicircles each have a diameter equal to the width \( w \), thus the radius of each semicircle is \( \frac{w}{2} \). The total perimeter involving the lengths and the semicircles sums to 500 ft.

Since there are two semicircles, their combined perimeter forms a full circle:

\[ \pi \times \frac{w}{2} + \pi \times \frac{w}{2} = \pi w \]

The entire perimeter is:

\[ 2l + \pi w = 500 \]

Solve for \( l \):

\[ l = \frac{500 - \pi w}{2} \]

The area \( A \) of the rectangular portion is:

\[ A = l \times w = \left( \frac{500 - \pi w}{2} \right) \times w \]

To find the maximum area, take the derivative \( \frac{dA}{dw} \) and set it equal to 0:

\[ \frac{dA}{dw} = \frac{d}{dw} \left( \frac{500w - \pi w^2}{2} \right) = \frac{500 - 2\pi w}{2} \]

Set the derivative to zero:

\[ \frac{500 - 2\pi w}{2} = 0 \]

\[ 500 - 2\pi w = 0 \]

\[ 2\pi w = 500 \]

\[ w = \frac{500}{2\pi} \]

Substitute this \( w \) back into \( l = \frac{500 - \pi w}{2} \):

\[ l = \frac{500 - \pi \left(\frac{500}{2\pi}\right)}{2} = \frac{500 - 250}{2} = 125 \]

Thus, the optimal length of the rectangular portion must be 125 feet.