Ask Question
20 February, 16:44

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?

+2
Answers (1)
  1. 20 February, 18:38
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Mathematics if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers