MATH SOLVE

2 months ago

Q:
# 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?

Accepted Solution

A:

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

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