Area of the floor
$A_1 = \frac{1}{4}\pi D^2$
$A_1 = \frac{1}{4}\pi (\frac{5}{6}H_1)^2$
$A_1 = \frac{25}{144}\pi {H_1}^2$
Area of cylindrical wall
$A_2 = \pi D H_2$
$A_2 = \pi (\frac{5}{6}H_1) \, H_2$
$A_2 = \frac{5}{6}\pi H_1 \, H_2$
Area of conical roof:
$A_3 = \pi r L$
$r = \frac{1}{2}D = \frac{1}{2}(\frac{5}{6}H_1) = \frac{5}{12}H_1$
$L = \sqrt{r^2 + {H_1}^2} = \sqrt{(\frac{5}{12}H_1)^2 + {H_1}^2} = \frac{13}{12}H_1$
Thus,
$A_3 = \pi (\frac{5}{12}H_1) (\frac{13}{12}H_1)$
$A_3 = \frac{65}{144}\pi {H_1}^2$
Total area:
$A = A_1 + A_2 + A_3$
$A = \frac{25}{144}\pi {H_1}^2 + \frac{5}{6}\pi H_1 \, H_2 + \frac{65}{144}\pi {H_1}^2$
$A = \frac{5}{8}\pi {H_1}^2 + \frac{5}{6}\pi H_1 \, H_2$
$\dfrac{dA}{dH_1} = \frac{5}{4}\pi H_1 + \frac{5}{6}\pi \left( H_1 \cdot \dfrac{dH_2}{dH_1} + H_2 \right) = 0$
$3 H_1 + 2 H_1 \cdot \dfrac{dH_2}{dH_1} + 2 H_2 = 0$
$\dfrac{dH_2}{dH_1} = -\dfrac{3 H_1 + 2 H_2}{2 H_1}$ → Equation (1)
Volume = volume of cylinder + volume of cone
$V = \frac{1}{4}\pi D^2 H_2 + \frac{1}{3} (\frac{1}{4}\pi D^2) H_1$
$V = \frac{1}{4}\pi (\frac{5}{6}H_1)^2 \, H_2 + \frac{1}{12}\pi (\frac{5}{6}H_1)^2 \, H_1$
$V = \frac{25}{144}\pi {H_1}^2 \, H_2 + \frac{25}{432}\pi {H_1}^3$
$\dfrac{dV}{dH_1} = \frac{25}{144}\pi \left( {H_1}^2 \cdot \dfrac{dH_2}{dH_1} + 2H_1 \, H_2 \right) + \frac{25}{144}\pi {H_1}^2 = 0$
${H_1}^2 \cdot \dfrac{dH_2}{dH_1} + 2H_1 \, H_2 + {H_1}^2 = 0$
$H_1 \cdot \dfrac{dH_2}{dH_1} + 2 H_2 + H_1 = 0$
$\dfrac{dH_2}{dH_1} = -\dfrac{H_1 + 2 H_2}{H_1}$ → Equation (2)
Equate Equations (1) and (2)
$\dfrac{dH_2}{dH_1} = \dfrac{dH_2}{dH_1}$
$-\dfrac{3 H_1 + 2 H_2}{2 H_1} = -\dfrac{H_1 + 2 H_2}{H_1}$
$2H_2 + 3H_1 = 4H_2 + 2H_1$
$H_1 = 2H_2$
Height of cone = 2 × height of cylinder answer