Consider a cuboid of sides 2 x , 4 x and 5 x and a closed hemisphere of radius r. If the sum of their surface areas is a constant k, then the ratio x: r, for which the sum of their volumes is maximum, is :

Sum of Surface areas $=76x^2+3\pi r^2=$constant (K)

$$\begin{gathered} V=40x^3+\frac{2}{3}\pi r^3 \\ \left[76x^2+3\pi r^2=K\right] \\ {r}^2=\frac{K-76x^2}{3\pi } \\ r={\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{1}{2}} \\ V=40x^3+\frac{2}{3}\pi {\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{3}{2}} \\ \frac{dV}{dx}=120x^2+\frac{2}{3}\pi ·\frac{3}{2}{\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{1}{2}}·\left(\frac{-76\left(2x\right)}{3\pi }\right) \end{gathered}$$

Put 

$$\begin{gathered} \frac{dV}{dx}=0\Rightarrow 120x^2+\frac{2}{3}\pi ·\frac{3}{2}{\left(\frac{K-76x^2}{3\pi }\right)}^{\frac{1}{2}}·\left(\frac{-76\left(2x\right)}{3\pi }\right)=0 \\ \Rightarrow 120x^2=\frac{152x}{3}{\left(\frac{k-76x^2}{3\pi }\right)}^{\frac{1}{2}} \\ \Rightarrow \frac{45}{19}{x}^2=x{\left(\frac{k-76x^2}{3\pi }\right)}^{\frac{1}{2}};x\ne 0 \\ \Rightarrow \frac{45}{19}x={\left(\frac{k-76x^2}{3\pi }\right)}^{\frac{1}{2}}\Rightarrow {\left(\frac{45}{19}\right)}^2{x}^2=\frac{k-76x^2}{3\pi } \\ \Rightarrow {\left(\frac{45}{19}\right)}^2{x}^2=r^2\Rightarrow \frac{x^2}{r^2}={\left(\frac{19}{45}\right)}^2 \\ \Rightarrow \frac{x}{r}=\frac{19}{45} \end{gathered}$$