Answer
Local maximum value at $x=2$ for a square with width and length is 2 in.
Work Step by Step
Let $x$ and $y$ be the sides of a rectangle with $xy=8$ and perimeter $2x+2y=8$
Thus, we have $f(x)=x(4-x)$ and $f'(x)=4-2x$
Need to find the maximal value of $f(x)$ .
when $f'(x)=0$ then $x=2$
This implies that the local maximum is at $x=2$ when $x \gt 0$.
$x=2$ and $y=4-2=2$ , that is, for a square and perimeter $= (2)(2)=8 in$
Hence, Local maximum value at $x=2$ for a square with width and length is 2 in.
