---
title: "Minimum material closed cylindrical tank"
subject: "Machine Design II"
chapter: 2
key: "semester_7/machine_design_ii:optimum-cylindrical-tank-calculus"
source: https://www.bctengineer.com/q/semester_7/machine_design_ii/optimum-cylindrical-tank-calculus
site: BCT Engineering
---

# Minimum material closed cylindrical tank

Machine Design II · Chapter 2: Optimization Techniques

- Asked in: year not given, 6 marks
- Video: [Designing an Oil can Using Least amount of Materials](https://www.youtube.com/watch?v=roVKcsPxCdA) — Zach Cresswell
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/machine_design_ii/optimum-cylindrical-tank-calculus
- Whole chapter: https://www.bctengineer.com/md/semester_7/machine_design_ii/chapter-2

## Question

A closed cylindrical steel tank must hold 2 m³ of liquid. Using differential calculus, find the diameter and height that need the least sheet-metal area. Show that your solution is a minimum and give the minimum area.

## Short answer

Express area A=2πr²+2V/r with V=2 m³. Set dA/dr=0: r³=V/(2π)→r=0.6828 m, so D=h=1.366 m. Second derivative 4π+4V/r³>0 confirms minimum. Minimum area is 8.79 m².

## Answer

**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 $= 37.7 > 0$).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. How is the area written as a function of one variable for the closed tank?
   A. A = 2 pi r^2 + 2V/r, after using h = V/(pi r^2)
   B. A = pi r^2 h, minimised directly
   C. A = 2 pi r^2 + 2 pi r h, with h left as free as r
   D. A = pi r^2 + 2 pi r h
   Answer: A. The objective is the closed-tank area 2 pi r^2 + 2 pi r h, and the volume constraint eliminates h to give 2 pi r^2 + 2V/r. (Key point: Formulation)

2. Setting dA/dr = 0 gives which condition?
   A. r^3 = V/pi
   B. r^2 = V/2
   C. r^3 = V/(2 pi), so r = 0.683 m for V = 2 m^3
   D. r = V/(2 pi)
   Answer: C. dA/dr = 4 pi r - 2V/r^2 = 0 gives r^3 = V/(2 pi) = 0.318, so r = 0.6828 m. (Key point: Stationary point)

3. What is the relation between height and diameter for the least-area closed tank?
   A. h = 2D
   B. h = D/2
   C. h = r
   D. h = D (height equals diameter)
   Answer: D. At the optimum h = V/(pi r^2) = 2r, so the height equals the diameter, 1.366 m. (Key point: Optimum shape)

4. Why is the stationary point a minimum?
   A. d2A/dr2 = 4 pi + 4V/r^3 = 37.7 is positive
   B. dA/dr is positive
   C. A is negative there
   D. d2A/dr2 is zero
   Answer: A. A positive second derivative means the area curve is concave upward at that point, so it is a minimum (8.79 m^2). (Key point: Second-derivative test)

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