Question

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?

Answer 1

Maximum area is `450` square feet when dimensions of play area are `30'xx15'`, 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+2w` and this is `60`feet. In other words `l+2w=60` i.e. `w=(60-l)/2=30-l/2`

Area covered by this will be `lxx(30-l/2)=30l-l^2/2`

= `-1/2(l^2-60l)`

= `-1/2(l^2-60l+900)+450`

= `-1/2(l-30)^2+450`

It is apparent that as coefficient of `(l-30)^2` is `-1/2`,

`-1/2(i-30)^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'xx15'`, where`30` feet is along the side of the house.