Find the minimum distance between point $M(0,-2)$ and points $(x,y)$ such that: $y=\frac{16}{\sqrt{3}\,x^{3}}-2$ for $x>0$ .
I used the formula for distance between two points in a plane to get: $$d=\sqrt{x^{2}+\frac{256}{3x^{6}}}$$ And this is where I cannot come up with how to proceed. I tried calculus but the first derivative of $d(x)$ is fairly ugly expression... A few techniques on how to handle problems on maxima and minima with(out) using calculus would be really useful.
On
HINT:
Searching a minimum value for $d^2$. I would say - in point minimum for $d^2$ will be minimized too 'd'.
On
We give a non-calculus approach. It is in my opinion a fair bit harder than the calculus way.
We want to minimize $x^2+\frac{256}{3x^6}$, or equivalently $$\frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{256}{3x^6}.$$ By the arithmetic mean geometric mean inequality (AM/GM) we have $$\frac{1}{4}\left( \frac{x^2}{3}+\frac{x^2}{3}+\frac{x^2}{3}+\frac{256}{3x^6} \right)\ge \sqrt[4]{\frac{x^2}{3}\cdot \frac{x^2}{3}\cdot\frac{x^2}{3}\cdot \frac{256}{3x^6}}\tag{1}$$ with equality when all the terms on the left are equal. The right-hand side of (1) is $\frac{4}{3}$. Equality is achieved when $\frac{x^2}{3}=\frac{256}{3x^6}$, that is, when $x=\pm 2$.
We conclude that the square of the distance has minimum value $\frac{16}{3}$.
Hint:
Note that the value of $x$ that minimize $d$ minimimize also $d^2=x^2+\dfrac{2^8}{3x^6}=f$ and the derivative is $f'= 2x-\dfrac{2^9}{x^7}$.