Answer
Minimal perimeter is 16 in for a square with width and length is 4 in.
Work Step by Step
Let $x$ and $y$ be the sides of a rectangle with $xy=16$ and perimeter $=2x+2y$
Thus, we have $f(x)=2x+2(\dfrac{16}{x})$ and $f'(x)=2-(\dfrac{32}{x^2})$
Need to find the minimal value of $f(x)$ .
when $f'(x)=0$ then $x= \pm 4$
This implies that the only critical point is at $x=4$ when $x \gt 0$
$x=4$ and $y=\dfrac{16}{4}=4$ , that is, for a square and perimeter $= (4)(4)=16 in$
Hence, Minimal perimeter is 16 in for a square with width and length is 4 in.
