Answer
Local maximum value at $x=2$ for a square with wide side and long side of 2 in.
Work Step by Step
We have $f(x)=x(4-x)=4x-x^2$ and $f'(x)=4-2x$
Consider $x$ and $y$ to be the sides of the rectangle.
We will have to find the maximal value of $f(x)$ .
When $x=2$, this implies $f'(x)=0$. This corresponds to a 2x2 square (perimeter of 2*2+2*2=8).
Hence, the local maximum value is at $x=2$ for a square with wide side and long side of 2 in.
