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A homeowner has 60 feet of fencing material to enclose a rectangular area for his pets to play in. One side of the house will be used as a side of the play area. What deminsions should be used in order to maximize the play area?

User Climbon
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2 Answers

5 votes

Answer:

i just want to say that the answer above is correct 15 ft by 30 ft is the write answer

Explanation:

i just took my test and got it right

User Shonte
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4 votes
Let the length of the side that would be a house be l. Given that the total length of the fencing material is 60 and the other side is x, then:
2x+l=60
⇒l=60-2x
The area of the house will therefore be:
A=length*width
A=x(60-2x)
A=60x-2x²
differentiating A w.r.t x we get
dA/dx=60-4x

equating the above to zero and solving for x we get
60-4x=0
hence
4x=60
⇒x=60/4
x=15
this implies that one of the sides of the area is 15 ft and the other is 30 ft
User Ahmad Vatani
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8.1k points
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