# A homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. He will use one side of his house as a side of the play area. What dimensions should he use if he wants to maximize the play area?

Jun 11, 2017

Maximum area is $450$ square feet when dimensions of play area are $30 ' \times 15 '$, where$30$ feet is along the side of the house.

#### Explanation:

Let $l$ be the length along the side of the house and $w$ be the width.

Hence fencing required will be $l + w + w = l + 2 w$ and this is $60$feet. In other words $l + 2 w = 60$ i.e. $w = \frac{60 - l}{2} = 30 - \frac{l}{2}$

Area covered by this will be $l \times \left(30 - \frac{l}{2}\right) = 30 l - {l}^{2} / 2$

= $- \frac{1}{2} \left({l}^{2} - 60 l\right)$

= $- \frac{1}{2} \left({l}^{2} - 60 l + 900\right) + 450$

= $- \frac{1}{2} {\left(l - 30\right)}^{2} + 450$

It is apparent that as coefficient of ${\left(l - 30\right)}^{2}$ is $- \frac{1}{2}$,

$- \frac{1}{2} {\left(i - 30\right)}^{2}$ is alwaays negative, except that it is $0$ when $l = 30$ and hence maximum area at this level is $450$ square feet and dimensions of play area will be $30 ' \times 15 '$, where$30$ feet is along the side of the house.