Questions: A track and field playing area is in the shape of a rectangle with semicircles at each end. See the figure. The inside perimeter of the track is to be 1500 meters. What should the dimensions of the rectangle be so that the area of the rectangle is a maximum? The rectangle has a width of and a length of (Use a comma to separate answers as needed. Type integers or decimals rounded to the nearest hundredth as needed.)

Solution Steps

Let $w$ be the width and $l$ be the length of the rectangle. The perimeter of the track is given by the sum of the lengths of the two lengths of the rectangle and the circumferences of the two semicircles (which form a full circle). The radius of the semicircle is $w/2$. The perimeter is $2l + 2\pi(\frac{w}{2}) = 1500$, so $2l + \pi w = 1500$.

From the perimeter equation, we can isolate $l$: $2l = 1500 - \pi w$, so $l = \frac{1500 - \pi w}{2}$.

The area of the rectangle is $A = lw$. Substituting the expression for $l$ from Step 2, we get $A(w) = w(\frac{1500 - \pi w}{2}) = 750w - \frac{\pi}{2}w^2$.

To maximize the area, we take the derivative of $A(w)$ with respect to $w$ and set it equal to zero: $A'(w) = 750 - \pi w = 0$. Solving for $w$, we get $w = \frac{750}{\pi}$.

Substitute the value of $w$ back into the equation for $l$ from Step 2: $l = \frac{1500 - \pi (\frac{750}{\pi})}{2} = \frac{1500 - 750}{2} = \frac{750}{2} = 375$.

Final Answer: The rectangle has a width of 238.73 and a length of 375.