Question 173895
Perimeter:  {{{P=2L + 2W}}}.  Since the perimeter is 140 feet, {{{2L + 2W = 140}}}


Solving for L:  {{{2L = 140 - 2W}}}, {{{L = 70 - W}}}


Area:  {{{A(LW) = LW}}}, but we know from the perimeter formula that {{{L=70-W}}} so {{{A(W)=(70-W)W=70W-W^2}}}


Since A(W) is continuous and differentiable the first derivitive set to zero gives the value of the independent variable at a local extrema.


{{{(dA(W))/dW=70-2W}}}, {{{70-2W=0}}}, {{{W=35}}}


Since the first derivitive is also differentiable, the sign on the second derivitive at the extreme point will characterize the extreme as a maximum or minimum.


{{{(d^2A(W))/dW^2=-2}}} for all {{{W}}} in the domain of {{{A(W)}}}, so the extreme point is a maximum.


Therefore, a rectangle with perimeter 140 has a maximum area when the width is 35, which means that the length is also 35 and the rectangle is actually a square.