Question 114605
A farmer has $2400 to spend on two rectangular pastures as shown to the right.
 (the diagram is of 2 identical shaped rectangles attached by one of the length
 sides). The local contractor will build the fence at a cost of $6.25/m.
 What is the largest total area the farmer can have fenced for that price??
:
Find out how many meters of fence he can get for $2400:
2400/6.25 =  384 meters of fence
:
Using a modified perimeter equation:
2L + 3W = 384
2L = 384 - 3W
L = {{{384/2}}} - {{{3/2}}}W
or
L = (192 - 1.5W)
:
Area = L * W
Substitute (192-1.5W) for L
A = W(192-1.5W) 
A = -1.5w^2 + 192w
:
Max area occurs at the axis of symmetry. Find the axis of symmetry: (x=-b/2a)
w = {{{(-192)/(2*-1.5)}}}
W = {{{(-192)/(-3)}}}
W = +64 meters
:
Find L:
L = 192 - 1.5(64)
L = 192 - 96
L = 96 meters
:
Max area: 96 * 64 = 6144 sq meters for $2400
:
You can confirm this by finding the vertex: Substitute 64 for w:
A = -1.5(64^2) + 192(64)
A = -6144 + 12288
A = +6144