58 - 59 Maxima and minima: cylinder surmounted by hemisphere and cylinder surmounted by cone

Let
k = unit price of wall
2k = unit price of floor
3k = unit price of roof
 

Total cost:
$C = 2k \, ( \, \frac{1}{4}\pi D^2 \, ) + k \, \pi D(H - \frac{1}{2}D) + 3k \, ( \, \frac{1}{2}\pi D^2 \, )$

$C = \frac{3}{2}k \, \pi D^2 + k \, \pi DH$

$\dfrac{dC}{dD} = 3k \, \pi D + k \, \pi \left( D \cdot \dfrac{dH}{dD} + H \right) = 0$

$3D + D \cdot \dfrac{dH}{dD} + H = 0$

$\dfrac{dH}{dD} = -\dfrac{H + 3D}{D}$   →   Equation (1)
 

Volume of silo = volume of cylinder + volume of hemisphere:
$V = \frac{1}{4}\pi D^2(H - \frac{1}{2}D) + \frac{1}{12}\pi D^3$

$V = \frac{1}{4}\pi D^2 H - \frac{1}{24}\pi D^3$

$\dfrac{dV}{dD} = \frac{1}{4}\pi \left( D^2 \cdot \dfrac{dH}{dD} + 2DH \right) - \frac{1}{8}\pi D^2 = 0$

$2D \cdot \dfrac{dH}{dD} + 4H - D = 0$

$\dfrac{dH}{dD} = -\dfrac{D - 4H}{2D}$   →   Equation (2)
 

Equate Equations (1) and (2)
$\dfrac{dH}{dD} = \dfrac{dH}{dD}$

$-\dfrac{H + 3D}{D} = -\dfrac{D - 4H}{2D}$

$-2H - 6D = D - 4H$

$2H = 7D$

$D = \frac{2}{7}H$
 

Diameter = 2/7 × total height           answer

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