Minimising the Cost of a Cylindrical Container
An engineering student is designing a cylindrical tin for a local food producer. The tin must hold exactly 600 cm³. The curved surface is made from a standard-grade metal costing €0.03 per cm². The top and bottom discs are made from a thicker, food-safe metal costing €0.08 per cm². A fixed seam and rim process adds €0.50 to the cost of every tin, independent of its dimensions. The student must choose the radius r and height h, both in centimetres, to minimise the total material and processing cost. The production manager has warned that the radius cannot exceed 8 cm because of the filling machinery, and the height must be at least 3 cm for the labelling equipment. The student sets up a cost function C(r) and needs to justify that the resulting dimensions give a minimum rather than a maximum.
1. Express the total cost C as a function of the radius r alone, and state the domain of r that satisfies the machinery constraints.
Read the worked answer
Step 1: Relate h to r.
The volume is fixed:
( \pi r^2 h = 600 \Rightarrow h = \dfrac{600}{\pi r^2} ).
Step 2: Build the cost function.
Curved surface area = (2\pi r h), costing €0.03 per cm².
Top and bottom area = (2\pi r^2), costing €0.08 per cm².
Fixed cost = €0.50.
So
[
C(r)=0.03(2\pi r h)+0.08(2\pi r^2)+0.50.
]
Substitute (h=\dfrac{600}{\pi r^2}):
[
C(r)=0.03\left(2\pi r\cdot\frac{600}{\pi r^2}\right)+0.16\pi r^2+0.50
= \frac{36}{r}+0.16\pi r^2+0.50.
]
Step 3: Apply constraints.
(r>0), (r\le 8), and (h=\dfrac{600}{\pi r^2}\ge 3).
The height constraint gives
[
\frac{600}{\pi r^2}\ge 3 \Rightarrow r^2\le \frac{200}{\pi}\approx 63.66
\Rightarrow r\le \sqrt{\frac{200}{\pi}}\approx 7.98\text{ cm}.
]
This is stricter than (r\le 8). Hence the domain is
[
0<r\le \sqrt{\frac{200}{\pi}}\approx 7.98\text{ cm}.
]
2. Find the value of r that minimises C, and calculate the corresponding height and minimum cost. Show that this is a minimum.
Read the worked answer
Step 1: Differentiate.
[
C(r)=\frac{36}{r}+0.16\pi r^2+0.50,
\qquad
C'(r)=-\frac{36}{r^2}+0.32\pi r.
]
Step 2: Solve (C'(r)=0).
[
\frac{36}{r^2}=0.32\pi r
\Rightarrow 0.32\pi r^3=36
\Rightarrow r^3=\frac{36}{0.32\pi}\approx 35.81.
]
Thus
[
r\approx \sqrt[3]{35.81}\approx 3.30\text{ cm}.
]
This lies inside the domain (0<r\le 7.98).
Step 3: Confirm a minimum.
[
C''(r)=\frac{72}{r^3}+0.32\pi>0\quad\text{for }r>0.
]
So the stationary point is a local minimum, and because (C''(r)>0) throughout the domain, it is the global minimum on that domain.
Step 4: Find h and minimum cost.
[
h=\frac{600}{\pi(3.296)^2}\approx 17.58\text{ cm}.
]
[
C_{\min}=\frac{36}{3.296}+0.16\pi(3.296)^2+0.50
\approx 10.92+5.46+0.50\approx €16.88.
]
So the cheapest tin has radius about (3.30) cm, height about (17.58) cm, and costs about €16.88.
3. The manager suggests simply using r = 5 cm because it is a round number. Compare the cost at r = 5 with the minimum, and comment on whether the difference is practically significant.
Read the worked answer
At (r=5) cm,
[
h=\frac{600}{\pi(5)^2}\approx 7.64\text{ cm},
]
and
[
C(5)=\frac{36}{5}+0.16\pi(5)^2+0.50
=7.20+12.57+0.50\approx €20.27.
]
The minimum cost was about €16.88, so using (r=5) cm costs
[
20.27-16.88\approx €3.39\text{ more per tin}.
]
As a percentage,
[
\frac{3.39}{16.88}\times 100\approx 20%.
]
A 20% increase is not trivial. For example, over 100,000 tins the extra cost would be about (100{,}000\times 3.39=€339{,}000). The difference is practically significant at production scale, so the optimised radius is worth using.