Minimum-Material Cans At Fixed Capacity

Why a Can Has an Optimal Proportion

A cylindrical can has two independent dimensions: its radius and its height. If its capacity is fixed, choosing a larger radius forces the can to be shorter; choosing a smaller radius forces it to be taller. These changes affect the amount of material needed for the curved wall and for the two circular ends.

We will model the can as a closed, right circular cylinder. The objective is to minimize its surface area while keeping its volume fixed. This model assumes uniform material, negligible thickness, and no material lost to seams, overlap, or cutting waste. Under these assumptions, less surface area means less material.

Design variables: Let \(r>0\) be the radius and \(h>0\) the height. For a closed cylinder, the surface area is the area of the curved wall plus the areas of both circular ends, and the volume is the capacity.

A(r,h)=2\pi r^2+2\pi rh, \qquad V(r,h)=\pi r^2h.

The term \(2\pi r^2\) accounts for the two ends, each of area \(\pi r^2\). The term \(2\pi rh\) is the curved wall: its area is the circumference \(2\pi r\) multiplied by height \(h\). We take the capacity \(V\) to be a specified positive constant.

Finding the Stationary Proportion

The design problem has two variables and one equality constraint. At a stationary constrained design, the gradient of the surface area is a scalar multiple of the gradient of the volume constraint. This is the Lagrange multiplier condition. Here the constraint is regular for every feasible can: its gradient is nonzero because \(r>0\).

Worked Example: Derive the stationary height-to-radius ratio

Write the constraint as \(g(r,h)=\pi r^2h-V=0\). The gradients are

\nabla A(r,h)=(4\pi r+2\pi h,\;2\pi r), \qquad \nabla g(r,h)=(2\pi rh,\;\pi r^2).

The Lagrange equations \(\nabla A=\lambda\nabla g\) are therefore

4\pi r+2\pi h=\lambda\,2\pi rh, \qquad 2\pi r=\lambda\,\pi r^2.

Since \(r>0\), the second equation gives \(\lambda=2/r\). Substituting this into the first equation gives

4\pi r+2\pi h = \frac{2}{r}(2\pi rh) = 4\pi h.

Subtracting \(2\pi h\) from each side yields \(4\pi r=2\pi h\), so \(h=2r\). The volume constraint then gives \(\pi r^2(2r)=V\), hence \(r^3=V/(2\pi)\). Thus the stationary dimensions are

r=\left(\frac{V}{2\pi}\right)^{1/3}, \qquad h=2\left(\frac{V}{2\pi}\right)^{1/3}.

To check the Lagrange equations, set \(h=2r\) and \(\lambda=2/r\). The first components satisfy \(4\pi r+2\pi h=8\pi r=\lambda(2\pi rh)\), and the second satisfy \(2\pi r=\lambda(\pi r^2)\). The volume is \(\pi r^2h=2\pi r^3=V\) for the radius just found.

The multiplier equations identify a stationary candidate; by themselves, they do not establish that the candidate is a minimum. We can prove global minimality by reducing the problem to one variable and checking the entire feasible range.

Why This Candidate Is the Global Minimum

For a fixed positive capacity \(V\), the constraint determines the height from any positive radius:

h=\frac{V}{\pi r^2}.

Substituting this expression into the area gives a function of the radius alone. Its domain is \(r>0\), since a can cannot have zero or negative radius.

A(r)=2\pi r^2+2\pi r\left(\frac{V}{\pi r^2}\right) =2\pi r^2+\frac{2V}{r}.

This reduction works because every positive radius corresponds to exactly one positive height satisfying the fixed-volume constraint. Consequently, minimizing this one-variable function is equivalent to minimizing area over all feasible cans.

Theorem: For every fixed capacity \(V>0\), the closed cylinder with least surface area has \(r=(V/(2\pi))^{1/3}\) and \(h=2r\). It is the unique global minimizer among cylinders with positive radius and height.

Proof. Differentiate the reduced area function for \(r>0\):

A'(r)=4\pi r-\frac{2V}{r^2} =\frac{2}{r^2}(2\pi r^3-V).

Because \(2/r^2\) is positive, the derivative is zero exactly when \(2\pi r^3=V\). There is exactly one positive solution, \(r_0=(V/(2\pi))^{1/3}\). If \(0<r<r_0\), then \(2\pi r^3<V\), so \(A'(r)<0\). If \(r>r_0\), then \(2\pi r^3>V\), so \(A'(r)>0\). The area therefore decreases up to \(r_0\) and increases after it.

This change in sign proves that \(r_0\) is the unique global minimum on the entire positive-radius domain. The corresponding height follows from the constraint:

h_0=\frac{V}{\pi r_0^2} =\frac{2\pi r_0^3}{\pi r_0^2} =2r_0.

Thus the minimizing can’s height is twice its radius. This completes the proof.

Calculating Dimensions for a Specified Capacity

Once the capacity is known, the result gives both dimensions directly. Keeping the units consistent matters: if volume is measured in cubic centimetres, the resulting radius and height are in centimetres, and the area is in square centimetres.

Worked Example: Design a can with capacity \(250\pi\) cubic centimetres

Let \(V=250\pi\text{ cm}^3\). The optimal radius is

r=\left(\frac{250\pi\text{ cm}^3}{2\pi}\right)^{1/3} =(125\text{ cm}^3)^{1/3} =5\text{ cm}.

The optimal height is twice the radius, so \(h=10\text{ cm}\). Check the capacity by substituting the dimensions into the volume formula:

V=\pi(5\text{ cm})^2(10\text{ cm}) =250\pi\text{ cm}^3.

The material area is

A=2\pi(5\text{ cm})^2+2\pi(5\text{ cm})(10\text{ cm}) =50\pi\text{ cm}^2+100\pi\text{ cm}^2 =150\pi\text{ cm}^2.

The dimensions reproduce the required capacity, and the area includes both ends as well as the curved wall.

At the optimum, the area can also be written in terms of the radius alone. Substituting \(h=2r\) into the full surface-area formula gives \(A=6\pi r^2\). This is a convenient check, but it is valid for the minimum design; it is not the surface area formula for an arbitrary cylinder.

Comparing a Feasible Design with the Minimum

A feasible design has the required capacity, but that alone does not mean it uses the least material. To assess a proposed design, first check its volume, then calculate its full surface area and compare it with the minimum for that capacity.

Worked Example: Compare two cans with the same capacity

Keep the capacity \(250\pi\text{ cm}^3\). Consider a can with \(r=10\text{ cm}\) and \(h=2.5\text{ cm}\). Its volume is

V=\pi(10\text{ cm})^2(2.5\text{ cm}) =250\pi\text{ cm}^3,

so it is feasible. Its surface area is

A=2\pi(10\text{ cm})^2+2\pi(10\text{ cm})(2.5\text{ cm}) =200\pi\text{ cm}^2+50\pi\text{ cm}^2 =250\pi\text{ cm}^2.

The minimum-area can for this capacity has \(r=5\text{ cm}\), \(h=10\text{ cm}\), and area \(150\pi\text{ cm}^2\). The proposed can uses \(250\pi-150\pi=100\pi\text{ cm}^2\) more area. The comparison is between two designs with exactly the same volume, so the difference comes from their proportions rather than their capacities.

A Common Pitfall: Forgetting One End

A closed can has two circular ends. Omitting one of them changes the mathematical model and produces a different, incorrect design for a closed can. For example, using \(\pi r^2+2\pi rh\) counts just one end. After substituting the fixed-volume constraint, this mistaken objective becomes \(\pi r^2+2V/r\), whose derivative vanishes when \(r^3=V/\pi\). The resulting stationary condition is \(h=r\), not \(h=2r\).

For \(V=250\pi\text{ cm}^3\), that incorrect condition would give \(r=h=\sqrt[3]{250}\text{ cm}\). It does satisfy the volume constraint, since \(\pi r^3=250\pi\text{ cm}^3\), but it is not the minimum-area closed can. The mistake is not in differentiating the one-end formula; it is in applying that formula to a can that is supposed to be closed at both ends. Restoring the missing disk gives the correct objective \(2\pi r^2+2\pi rh\), and the correct optimal ratio \(h=2r\).

Model check: Before optimizing, identify every surface included in the design. The ratio \(h=2r\) applies to a closed cylinder with both ends made of the same material per unit area as the wall. Changing those material costs changes the model.

Check Your Understanding

Use the fixed-volume model and the global-minimum argument to answer these questions.

Write the surface-area formula for a closed cylinder and identify which terms represent the ends and the curved wall.
For fixed positive capacity, express \(h\) in terms of \(V\) and \(r\), and use it to write the area as a function of \(r\) alone.
Why does the sign of the derivative on either side of the critical radius establish a global minimum?
A can has capacity \(128\pi\text{ cm}^3\). Find its minimum-area radius and height.
Explain how forgetting one circular end changes the stationary height-to-radius ratio.