Problema Solution
A rectangle of length "x" and width "w" has a perimeter of 72 units. Give the area equation in terms of "x" only, graph that equation, and then give the dimensions of the rectangle that yields maximum area.
Answer provided by our tutors
area = l*w
perimeter = 2(l+w) where 'l' and 'w' are length and width, respectively
72 = 2(x+w) = 2x + 2w
2 w = 72 - 2x
w = (72-2x)/2 = -x + 36
let 'A' represent area, then:
A = x*w
substitute '(-x+36)' for 'w':
A = x(-x+36) = -x^2 + 36x
Area in terms of x, then is:
A = -x^2 + 36x = -(x-18)^2 + 324
to maximize area, the we want to minimize the always-negative term "-(x-18)^2" and choosing x=18 sets this term to zero (this will also mark the apex of the parabola when graphed)
the dimensions that yield the maximum area are
width = 18
72 = 2(x+18)
solving for 'x' we have x=18
the maximum area is a square of 18x18 units