Question 176361
Since 1 side is the barn, the farmer needs only 3 sides.There will
be 2 equal length sides perpendicular to the barn and one side parallel 
to the barn. 
Let {{{x}}} = the length of a side perpendicular to barn
Let {{{z}}} = the length of the side parallel to the barn
Let {{{A}}} = the area enclosed by the fencing
Given:
{{{100 = 2x + z}}}
{{{z = 100 - 2x}}}
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{{{A = xz}}} formula for the area of a rectangle
And, from above,
{{{A = x*(100 - 2x)}}}
{{{A = 100x - 2x^2}}}
{{{A = -2x^2 + 100x}}}
This is a function {{{A(x)}}}. I can plot it and find the
maximum. I also know that, given a general equation
{{{y = ax^2 + bx + c}}}, there is a maximum if the sign of
{{{a}}} is negative. That maximum occurs at
{{{x = -(b/(2a))}}}. In this case, 
{{{a = -2}}}
{{{b = 100}}}
{{{-(b/(2a)) = -(100/(2*-2))}}}
{{{-(100/(-4)) = 25}}}
{{{x = 25}}} ft
When {{{x=25}}},
{{{z = 100 - 2x}}}
{{{z = 100 - 50}}}
{{{z = 50}}} ft
{{{A[max] = xz}}}
{{{A[max] = 25*50}}}
{{{A[max] = 1250}}} ft2
I'll plot the equation A(x), too
{{{ graph( 500, 500, -10, 100, -20, 1500, -2x^2 + 100x) }}}
A good way to check to see if you have a maximum is
to just tweak the numbers a tiny bit, like
{{{x = 24.8}}}
{{{z = 50.4}}} What's the area?
{{{A = 1249.92}}} The area went down as it should
Then tweak the other way
{{{x = 25.2}}}
{{{z = 49.6}}}
{{{A = 1249.92}}} area also went down
So, {{{A - 1250}}} looks like a maximum