How to calculate the local minimum of a hyperbola without using derivatives?

Question

I've got the following rational function, which is a hyperbola.

$f(x) = \frac{2\cdot\pi (x+ 4)^2}{x}$$\quad$on WolframAlpha

There is a minimum in the first quadrant and a maximum in the third quadrant. I want to find the minimum in the first quadrant, so I define that $x>0$.

Now I want to find this minimum without using the derivation of the function $f(x)$.

My idea was to remove somehow the part I don't want to look at (with $x>0$), so the function is not rational anymore. Then I would use the vertex of a parabola notation $\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)$. (What's the correct name for this in English?)

$\frac{2\cdot\pi (x+ 4)^2}{x} \Rightarrow 2\cdot\pi (x+ 4)^2 \Rightarrow 2\pi x^2 + 16\pi x + 32\pi$

This leads to $-\frac{16\pi}{4\pi}$, which leads to $x=-4$. Now where I'm struggling is to get the minimum, because this seems to be the maximum?

Answer 1

We want to minimize $\frac{2\pi(x+4)^2}{x}$ for positive $x$. Expanding, we want to minimize $$2\pi\left(x+8+\frac{16}{x}\right).$$ Equivalently, we minimize $$\left(x+\frac{16}{x}\right)^2.$$ This is equal to $$64+\left(x-\frac{16}{x}\right)^2.$$ The above expression is minimized when the second term is $0$. This occurs when $x=4$.

Answer 2

Since $x>0$, from AM-GM, we have $$\dfrac{(x+4)^2}x = x + \dfrac{16}x + 8 \geq 2 \sqrt{16}+8 = 16$$ And equality (or the minimum occurs) when $x = \frac{16}x$, i.e., when $x=4$.