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

**Problem 58**

For the silo of Problem 57, find the most economical proportions, if the floor is twice as expensive as the walls, per unit area, and the roof is three times as expensive as the walls, per unit area.

**Solution:**

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*

**Problem 59**

An oil can consists of a cylinder surmounted by a cone. If the diameter of the cone is five-sixths of its height, find the most economical proportions.

**Solution:**

$A_1 = \frac{1}{4}\pi D^2$

$A_1 = \frac{1}{4}\pi (\frac{5}{6}H_1)^2$

$A_1 = \frac{25}{144}\pi {H_1}^2$

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:

$A_3 = \pi r L$

$L = \sqrt{r^2 + {H_1}^2} = \sqrt{(\frac{5}{12}H_1)^2 + {H_1}^2} = \frac{13}{12}H_1$

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*

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