Question

Optimization Question Calculus : Please zoom in.

Answer 1

So hmm check the picture below, that's the silo, notice the top of the cylinder is inside with grain... and the bottom, well, that's the (base no included)

so

$\bf \begin{cases} V_c=\textit{volume of the cylinder}\\ V_h=\textit{volume of the hemisphere}\\ A_c=\textit{area of the cylinder}\\ A_h=\textit{area of the hemisphere} \end{cases}\\\\ -------------------------------$

$\bf \textit{total volume of silo }14000=\pi r^2h+\cfrac{2\pi r^3}{3}\implies 14000-\cfrac{2\pi r^3}{3}=\pi r^2h \\\\\\ \boxed{\cfrac{14000}{\pi r^2}-\cfrac{2r}{3}=h}\\\\ -------------------------------$

$\bf \textit{now onto the cost for each area}\\\\ A_c=2\pi r\left( \cfrac{14000}{\pi r^2}-\cfrac{2r}{3} \right)\implies A_c=\cfrac{28000}{r}-\cfrac{4\pi r^2}{3}\impliedby \begin{array}{llll} \textit{cost of 1}\\ stays\\ \textit{the same} \end{array} \\\\\\ A_h=2\pi r^2\impliedby \textit{cost is twice as much} \\\\\\ A_h=4\pi r^2\\\\ -------------------------------$

$\bf \textit{total cost}\qquad C(r)=\cfrac{28000}{r}-\cfrac{4\pi r^2}{3}+4\pi r^2 \\\\\\ C(r)=\cfrac{28000}{r}-\cfrac{4\pi r^2}{3}+\cfrac{12\pi r^2}{3}\implies C(r)=\cfrac{28000}{r}+\cfrac{8\pi r^2}{3} \\\\\\ \boxed{C(r)=28000r^{-1}+\cfrac{8\pi }{3}r^2}\\\\ -------------------------------$

$\bf \textit{now, let's check the critical points}\\\\ \cfrac{dc}{dr}=-\cfrac{28000}{r^2}+\cfrac{16\pi r}{3}\implies \cfrac{dc}{dr}=\cfrac{16\pi r^3-84000}{3r^2} \\\\\\ 0=\cfrac{16\pi r^3-84000}{3r^2}\implies 84000=16\pi r^3\implies 5250=\pi r^3 \\\\\\ \boxed{\sqrt[3]{\cfrac{5250}{\pi }}=r}$

now, I'd like to point out, we also get critical points from the denominator being zero, so 3r²=0, gives us r=0, which is a critical point, however, with a radius of 0, well, there's no volume, so, the is not feasible for the phenomena, so is not applicable, though is a valid critical point

critical points from the denominator being zero, are usually asymptotic or "cusps", the function reaches an extrema, however the tangent is not horizontal and thus is not differentiable at that point, even though is an extrema.