You have a 600-foot roll of fencing to enclose two separate rectangular playgrounds for different age groups. If the two playgrounds are identical and adjacent, and form a larger rectangular region, what is the largest possible area of the playgrounds in total?

Question

Quizplus · Accepted Answer

Let the length of each playground be $l$ and the width be $w$. Since the two playgrounds are identical and adjacent, the larger rectangle has length $2l$ and width $w$.

Since there are 600 feet of fencing, we have $2(2l+w)=600$, or $4l+2w=600$. Solving for $w$ in terms of $l$, we get $w=300-2l$, or $l=\frac{1}{2}(300-w)$.

The area of a rectangle is given by $lw$. Substituting for $l$ in terms of $w$, we have that the total area is $2\left(\frac{1}{2}(300-w)\right)w=300w-w^2$.

To find the maximum value of this expression, we complete the square: $$300w-w^2=-\left(w^2-300w \right)=-(w^2-300w+22500-22500)=-(w-150)^2+22500.$$

Since $(w-150)^2 \geq 0$ for all $w$, it follows that $-(w-150)^2 \leq 0$, and hence the maximum value of $300w-w^2$ is $22,\!500$, which occurs when $w=150$.

Plugging this value of $w$ back in, we can solve for the length of each playground: $l=\frac{1}{2}(300-w)=75$. Therefore, the total area of the two playgrounds is $2lw=2(75)(150)=\boxed{22,\!500}$ square feet.

You Have a 600-Foot Roll of Fencing to Enclose Two