$S = x^2/2 + (16000√2)/x$
I know I have to find the derivative and get the stationary points, then use the second derivative to find the one which gives a minimum value and then put that x value back into S to find the minimum value.
However, it seems like there are some issues with how I am differentiating because I keep getting absurd answers which I won't waste time outlining..
any advice?
Cheers
Best Answer
Let $f(x)=\frac{1}{2}x^2+\frac{16000\sqrt{2}}{x}$. We have that $$ f'(x)=\frac{1}{2}(2x)+\frac{-16000\sqrt{2}}{x^2}=x-\frac{16000\sqrt{2}}{x^2}. $$ To find the extrema, we set $f'(x)=0$, so $$ \begin{align*} x-\frac{16000\sqrt{2}}{x^2}=0 &\iff x=\frac{16000\sqrt{2}}{x^2} \\ &\iff x^3=16000\sqrt{2} \\ &\iff x=20\sqrt{2}. \end{align*} $$ Now, we find $f''(20\sqrt{2})$, $$ f''(x)=1+\frac{32000\sqrt{2}}{x^3} \ \ \ \textrm{so} \ \ \ f''(20\sqrt{2})=3. $$ Since $f''(20\sqrt{2})>0$, we know the function is concave up at $x=20\sqrt{2}$, thus $f(20\sqrt{2})=1200$ is a minimum of the function. If you look at the graph of the function though, it is pretty clear that the function only has a local minimum, not an absolute one, that is if you are considering the domain of $f$ to be $\mathbb{R}$ and not $\mathbb{R}^+$.