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?
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
