Question 37629
the fields would look like 3 adjoining rectangles
with $3600 to spend and and a cost of $6.25/m, the farmer can
afford 3600/6.25 = 576 m of fence
He needs 4 north-south lengths and 2 east-west lengths (for example)
in order to enclose 3 rectangular fields
All these lengths will add up to 576 m.
{{{2x + 4y = 576}}}
The area enclosed is x times y
{{{A = xy}}}
solve for y in the 1st equation
{{{4y = 576 - 2x}}}
{{{y = 144 - x/2}}}
substitute this in equation 2
{{{A = x(144 - x/2)}}}
{{{A = -x^2/2 + 144x}}}
graph the equation
{{{ graph( 300, 500, -50, 300, -500, 12000, -x^2/2 + 144*x) }}}
I'll guess the x value where the graph is a max is 144
{{{A = -x^2/2 + 144x}}}
{{{A = -(144)^2/2 + 144*(144)}}}
factor out 144^2
{{{A = 144^2*(1 - 1/2)}}}
{{{A = 144^2 / 2}}}
{{{A = 20736/2}}}
{{{A = 10368}}}
The way to test this is to increase 144 a little, test it in the equation
then decrease 144 a little and test it in the equation
{{{A = -(143)^2/2 + 144*(143)}}}
{{{A = 10367.5}}}
{{{A = -(145)^2/2 + 144*(145)}}}
{{{A = 10367.5}}}
since the area is a little bit less than 10368 on either side
of x = 144, then this is the max.
The largest Area that can be enclosed is 10368
----------------------------
check
If x = 144, find y
xy = 10368
y = 10368/144
y = 72
{{{2x + 4y = 576}}}
{{{2*144 + 4*72 = 576}}}
{{{288 + 288 = 576}}}
{{{576 = 576}}}
OK