<PRE><B>Question 852390</B>
A rectangular area is enclosed by a fence and divided by another section of fence parallel to two of its side.
 The 500m of fence used to enclose a maximum area.
 What are the dimensions of the fence? (leave answers in fraction form)
:
Because of the divided area, the perimeter will be:
2L + 3W = 500
therefore
2L = -3W + 500
L = {{{-3/2}}}W + {{{500/2}}}
L = -1.5W + 250
:
Area = L*W
Replace L with (-1.5W+250)
A = (-1.5W+250)*W
A = -1.5W^2 + 250W
A quadratic equation, the max area occurs on the axis of symmetry x=-b/(2a)
In this equation, x=w, a=-1.5, b=250
w = {{{(-250)/(2*-1.5)}}}
W = {{{(-250)/(-3)}}}
W = +83{{{1/3}}} meters is with width that gives max area
Find the Length
L = -1.5w+250
L = -1.5(83.33) + 250
L = -125 + 250
L = 125 meters is length for max area
Then
125 * 83{{{1/3}}} = 10416{{{2/3}}} sq meters is the max area</PRE>