# A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. 548 feet of fencing is used. How do you find the dimensions of the playground that maximize the total enclosed area?

Oct 23, 2015

$137 \times 91 \frac{1}{3}$ feet

#### Explanation:

Let
$\textcolor{w h i t e}{\text{XXX}} L =$ length of entire playground
$\textcolor{w h i t e}{\text{XXX}} W =$ width of entire playground.

Assume the dividing fence is parallel to the width.

Total fencing required
$\textcolor{w h i t e}{\text{XXX}} 2 L + 3 W = 548$

$\textcolor{w h i t e}{\text{XXX}} L = \frac{548 - 3 W}{2}$

Area of entire playground
$\textcolor{w h i t e}{\text{XXX}} A = L \times W$

$\textcolor{w h i t e}{\text{XXXX}} = \frac{548 - 3 W}{2} \cdot W$

$\textcolor{w h i t e}{\text{XXXX}} = - \frac{3}{2} {W}^{2} + 274 W$

Maximum area will occur at a point where the derivative of the area is equal to zero.

$\textcolor{w h i t e}{\text{XXX}} \frac{\mathrm{dA}}{\mathrm{dW}} = - 3 W + 274 = 0$

$\textcolor{w h i t e}{\text{XXX}} W = 91 \frac{1}{3}$

Substituting $91 \frac{1}{3}$ for $W$ in $2 L + 3 W = 548$
gives
$\textcolor{w h i t e}{\text{XXX}} L = 137$