Ahomeowner 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?

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

(answer 2 )i’m very smart that’s how i know the answer.

X=y+4 is the answe