Minimum material closed cylindrical tank

Method: optimisation by differential calculus. Write the objective (surface area), eliminate one variable using the constraint (volume), set the first derivative to zero and check the second derivative.

Formulation

Let $r$ = radius and $h$ = height.

• Objective: $A = 2\pi r^2 + 2\pi r h$ (minimise)
• Constraint: $V = \pi r^2 h = 2\ \text{m}^3 \Rightarrow h = \dfrac{V}{\pi r^2}$

Substitute $h$:

$$A(r) = 2\pi r^2 + \frac{2V}{r}$$

Stationary point

$$\frac{dA}{dr} = 4\pi r - \frac{2V}{r^2} = 0 \;\Rightarrow\; r^3 = \frac{V}{2\pi}$$ $$r = \left(\frac{2}{2\pi}\right)^{1/3} = 0.6828\ \text{m}$$

Then $D = 2r = 1.3656$ m and

$$h = \frac{V}{\pi r^2} = \frac{2}{\pi (0.6828)^2} = 1.3656\ \text{m} = 2r$$

so the best tank has height equal to diameter.

Check for minimum

$$\frac{d^2A}{dr^2} = 4\pi + \frac{4V}{r^3} = 12.566 + \frac{8}{0.3183} = 37.7 > 0$$

The second derivative is positive, so this is a minimum.

Minimum area

$$A_{min} = 2\pi (0.6828)^2 + 2\pi (0.6828)(1.3656) = 2.929 + 5.859 = 8.79\ \text{m}^2$$

Answer: $D = h = 1.366$ m, $A_{min} = 8.79\ \text{m}^2$ (second derivative $= 37.7 > 0$).