The back of George's property is a creek. George would like to enclose a rectangular area, using the creek as one side and fencing for the other three sides, to create a pasture. If there is 460 feet of fencing available, what is the maximum possible area of the pasture?

Answer 1

Let the side parallel to the creek be y ft
let the other two sides be x ft each
so you have y + 2x = 460
y = 460-2x

area = xy = x(460-2x)
= -2x^2 + 460x

this can be represented by a downwards pointing parabola of the form y = ax^2+bx+c
the x of such a parabola is -b/(2a)
= -460/-4 = 115
if x = 115, then y = 460-2(115) = 230

max area = xy = 115(230) = 26450 ft^2

or , by Calculus
d(area)/dx = -4x + 460 = 0 for a max of area
4x = 460
x = 115 , just as before, continue....