Question

Optimization Calculus Problem

Answer 1

$\bf \begin{cases} V_c=\textit{cylinder's volume}\\ V_h=\textit{hemisphere's volume}\\ A_c=\textit{area of the cylinder}\\ A_h=\textit{area of the hemisphere} \end{cases} \\\\\\ V_c+V_h=6000\implies (\pi r^2h)+\left( \frac{2\pi r^3}{3} \right) \\\\\\ 6000=\cfrac{3\pi r^2h+2\pi r^3}{3}\implies 18000-2\pi r^3=3\pi r^2h \\\\\\ \boxed{\cfrac{6000}{\pi r^2}-\cfrac{2r}{3}=h} \\\\ -------------------------------$



$\bf A_c=2\pi rh+2\pi r^2\implies A_c=2\pi r\left(\cfrac{6000}{\pi r^2}-\cfrac{2r}{3} \right)+2\pi r^2 \\\\\\ A_c=\cfrac{12000}{r}-\cfrac{4\pi r^2}{3}\impliedby \textit{say it has a cost of 1, so stays the same}\\\\ -------------------------------\\\\ A_h=2\pi r^2\impliedby \textit{has a cost of 8 times the cylinder's} \\\\\\ A_h=16\pi r^2\\\\ -------------------------------$


$\bf \textit{total cost is }A_c+A_h \\\\\\ \cfrac{12000}{r}-\cfrac{4\pi r^2}{3}+16\pi r^2\implies \cfrac{12000}{r}-\cfrac{4\pi r^2}{3}+\cfrac{48\pi r^2}{3} \\\\\\ \boxed{C(r)=\cfrac{12000}{r}+\cfrac{44\pi r^2}{3}}$