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*