Problems

Solution: \(V_h = \frac43 \pi r_h^3, S_h = 4 \pi r_h^2\) \(\frac{\d V_h}{\d t} = -k4\pi r_h^2 \Rightarrow 4\pi r_h^2 \frac{\d r_h}{\d t} = -k 4\pi r_h^2 \Rightarrow \frac{\d r_h}{\d t} = -k\) Therefore \(r_h = 2R - kt, r_b = 3R - kt\). The height will halve when \(2kt = \frac{5}{2}R \Rightarrow kt = \frac{5}{4}R\) and the two sections will have radii \(\frac{3}{4}R\) and \(\frac{7}{4}R\) and the ratio of the volumes will be: \begin{align*} \frac{\frac{3^3}{4^3}+\frac{7^3}{4^3}}{2^3+3^3} = \frac{37}{224} \end{align*} \begin{align*} && \frac{\d V}{\d t} &= -4\pi k(r_h^2+r_b^2) \\ && \frac{\d S}{\d t} &= -8\pi k (r_h+r_b) \\ \Rightarrow && \frac{\d V}{\d S} &= \frac{r_h^2 + r_b^2}{2(r_h+r_b)} \\ &&&= \frac{(2R-kt)^2+(3R-kt)^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rkt+2k^2t^2}{2(5R-2kt)} \\ &&&= \frac{13R^2-10Rs + 2s^2}{2(5R-2s)} \end{align*} Where \(s = kt\) and \(0 \leq s \leq 2R\). We can maximise this but differentiating wrt to \(s\). \begin{align*} \Rightarrow && &= \frac{(-10R+4s)(10R-4s)+4(13R^2-10Rs+2s^2)}{4(5R-2s)^2} \\ &&&= \frac{-48R^2+40Rs-8s^2}{4(5R-2s)^2} \\ &&&= \frac{-8(s-2R)(s-3R)}{4(5R-2s)^2} \\ &&&<0 \end{align*} Therefore it is largest when \(s = 0\), ie \(\frac{13R^2}{10R} = \frac{13}{10}R\)