# 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 want to maximize the play area.

Attached is an answer using algebra.

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The homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. One side of the play area would be his house. The dimensions of the play area have to be determined to maximize it.

Let the length of the side that would be the house be L. As the total length of fencing material is 60, if the other side is x, 2x + L = 60

=> L = 60 - 2x

The area of the house is A = x*(60 - 2x) = 60x - 2x^2

Differentiate A with respect to x, equate it to 0 and solve the equation for x.

`(dA)/(dx) = 60 - 4x`

60 - 4x = 0

=> x = 15

One of the sides of the area is 15 and the other is 30 feet.

To maximize the area, the dimensions of the play area are 15 feet and 30 feet.