Answer
Minimal perimeter is 16 in for a square whose length side and width side is $4 in$.
Work Step by Step
Since, we have $f(x)=2x+2(\dfrac{16}{x})$ and also, $f'(x)=2-(\dfrac{32}{x^2})$
Let $x$ and $y$ be the sides of a rectangle with area $xy=16$ and perimeter $P=2x+2y$
We will have to find the minimal value of $f(x)$ .
when $f'(x)=0$ this implies $x= \pm 4$
Here, the only critical point is at $x=4$ (when $x \gt 0$)
Thus, we have $xy=16$
Therefore, $x=4$ and $y=(\dfrac{16}{4})=4$ ,
These are dimensions for a square with perimeter $P= (4)(4)=16 in$
Hence, the minimal perimeter is 16 in for a square whose length side and width side is $4 in$.
