Problema Solution
A rectangular playground is to be fenced off and divided in two by another fence parallel to one side of the playground. Four hundred and eighty feet of fencing is used. Find the dimensions of the playground that maximize the total enclosed area.
What is the maximum area?
What is the smaller dimension?
What is the Larger dimension?
Answer provided by our tutors
let 'x' and 'y' be the sides of the rectangular playground
the total length of the fence used is 480 ft
2x + 2y + y = 480
2x + 3y = 480
y = 160 - (2/3)x
A = x*y
plug y = 160 - (2/3)x into the last equation
A = x(60 - (2/3)x)
A = -(2/3)x^2 + 60x
we need to find the maximum
A max = c - b^2/(4a)
A max = - 60^2/(4*(-2/3))
A max = 1350 is the maximum area
x = -b/2a
x = - 60/(2*(-2/3))
x = 45 ft is the larger dimension
y = A/x
y = 1350/45
y = 30 ft is the smaller dimension