Question 112897
If the side parallel to the barn is x and the area is 2050, then the two sides perpendicular to the barn are each {{{2050/x}}}.


Let L(x) be the length function and {{{L(x) = x}}}.
Let W(x) be the width function and {{{W(x)=2050/x}}}
Let P(x) be the perimeter function representing the length of the three fenced sides, hence {{{P(x)=L(x)+2W(x)}}}


P(x) will have a critical point where {{{dP(x)/dx = 0}}}.


{{{dP(x)/dx=dL(x)/dx+2dW(x)/dx}}}


{{{dP(x)/dx=1-2*2050/x^2}}}


{{{1-2*2050/x^2=0}}}
{{{x^2-4100=0}}}
{{{x=sqrt(4100)}}}


Now the question is whether this critical point is a minimum or a maximum.


If {{{d^2P(sqrt(4100))/dx^2>0}}} then the critical point is a minimum.


{{{d^2P(x)/dx^2=2*4100/x^3}}}


{{{d^2P(sqrt(4100))/dx^2=2*4100/(sqrt(4100))^3>0}}}, therefore the critical point is a minimum.


So {{{sqrt(4100)}}} is the length x that gives the minimum length fence.  The sides are then {{{2050/sqrt(4100)}}} or {{{sqrt(4100)/2}}} when you rationalize the denominator.  I would submit the answer in this form since these values are exact.  If you want to do the calculator work, you will get x to be a little larger than 64 (since {{{64^2=4096}}} and the sides will be a little more than 32, giving you a bit more than 128 feet of fencing required.


Hope this helps,
John