Introduction
A cylindrical can must hold a specified amount, but its dimensions are not determined by capacity alone. A can with a small radius and great height can hold the same volume as a wide, shallow can. Their surface areas differ, so the amount of material needed differs as well.
We will treat the can as a closed right circular cylinder: it has a curved side and two circular ends. The goal is to choose its radius and height to minimize the total surface area while keeping its volume fixed.
This is an idealized material model. It assumes the can has uniform thickness and counts the area of its surfaces, but not extra material for seams, overlaps, rims, or manufacturing waste. Under these assumptions, minimizing surface area minimizes the amount of material.
Set Up the Constraint and Objective
Let \(r\) be the cylinder’s radius and \(h\) its height. Both dimensions must be positive. Let the prescribed volume be \(V\), where \(V>0\). The cylinder’s volume is its circular base area multiplied by its height.
The material area has two parts. The top and bottom are two circles, each with area \(\pi r^2\). The curved side can be opened into a rectangle with width equal to the circumference \(2\pi r\) and height \(h\), so its area is \(2\pi rh\). Therefore the total area is
The volume condition lets us eliminate one variable. Solving \(V=\pi r^2h\) for \(h\), for \(r>0\), gives
Substitute this expression into the area formula. The resulting objective depends only on \(r\):
The first term counts the two circular ends; the second counts the curved side. If \(r\) is measured in units of length and \(V\) in cubic units, then both terms have units of area: \(r^2\) is an area, and \(V/r\) is also an area.
Find the Candidate Dimensions
Differentiate the reduced area function for \(r>0\):
A critical point occurs where \(A'(r)=0\). Multiplying that equation by \(r^2\), which is positive on the feasible domain, gives
Thus the candidate radius is \(r=(V/(2\pi))^{1/3}\). Substituting the critical-point condition into the volume constraint gives the matching height:
The dimensions at the critical point therefore satisfy a simple geometric relation: the height equals the diameter. This relation is a consequence of minimizing the area with volume fixed; it is not an extra assumption about the can.
Why This Candidate Is the Global Minimum
A critical point alone does not establish that an objective has a global minimum. Here we can determine the sign of the derivative throughout the feasible domain. Rewrite it as
The factor \(2/r^2\) is positive for every feasible radius. The sign of \(A'(r)\) is therefore the sign of \(2\pi r^3-V\). Define \(r_*=(V/(2\pi))^{1/3}\), the positive radius that satisfies \(2\pi r_*^3=V\). Since \(r^3\) increases as positive \(r\) increases, \(A'(r)<0\) when \(0<r<r_*\), and \(A'(r)>0\) when \(r>r_*\). The area decreases up to \(r_*\) and increases after it.
Because \(2\pi r^3<V\), the derivative is negative and increasing the radius reduces the area.
The derivative is zero, and the volume constraint gives the unique corresponding height \(h=2r_*\).
Because \(2\pi r^3>V\), the derivative is positive and further increasing the radius increases the area.
These signs show that every feasible radius other than \(r_*\) gives a larger area. Thus this is a global minimum on \(r>0\), not just a local minimum. The behavior at the ends of the domain agrees: as \(r\) approaches zero, \(2V/r\) grows without bound; as \(r\) grows without bound, \(2\pi r^2\) grows without bound.
To obtain the stated minimum area, substitute \(h_*=2r_*\) into the original surface-area formula: \(A=2\pi r_*^2+2\pi r_*(2r_*)=6\pi r_*^2\). The expression has units of area, as required.
Worked Examples
Worked Example: A Can with Volume \(54\pi\) Cubic Centimeters
Find the minimum-material dimensions and area for a closed can with volume \(V=54\pi\text{ cm}^3\).
The critical-radius condition is \(2\pi r^3=V\). Substitute the given volume:
The optimal height is twice the radius, so \(h=2(3\text{ cm})=6\text{ cm}\). Check the volume using these dimensions:
which is the required volume. The two circular ends have total area \(2\pi(3\text{ cm})^2=18\pi\text{ cm}^2\). The side has area \(2\pi(3\text{ cm})(6\text{ cm})=36\pi\text{ cm}^2\). Thus
Checking the reduced objective gives the same result: \(A(3\text{ cm})=2\pi(3\text{ cm})^2+2(54\pi\text{ cm}^3)/(3\text{ cm})=18\pi\text{ cm}^2+36\pi\text{ cm}^2=54\pi\text{ cm}^2\). The critical condition also checks: \(2\pi(3\text{ cm})^3=54\pi\text{ cm}^3=V\).
Worked Example: A Can with Volume \(16\pi\) Cubic Meters
Find the optimal dimensions and minimum surface area for a closed can with volume \(V=16\pi\text{ m}^3\).
Use the critical-radius condition and solve for \(r\):
Then \(h=2r=4\text{ m}\). Verify the capacity:
The ends require \(2\pi(2\text{ m})^2=8\pi\text{ m}^2\) of material, and the side requires \(2\pi(2\text{ m})(4\text{ m})=16\pi\text{ m}^2\). Therefore the total is \(8\pi\text{ m}^2+16\pi\text{ m}^2=24\pi\text{ m}^2\). The reduced objective confirms the calculation: \(A(2\text{ m})=2\pi(2\text{ m})^2+2(16\pi\text{ m}^3)/(2\text{ m})=8\pi\text{ m}^2+16\pi\text{ m}^2=24\pi\text{ m}^2\). Finally, \(2\pi r^3=2\pi(2\text{ m})^3=16\pi\text{ m}^3\), so the critical condition holds.
Worked Example: Compare an Alternative Can with the Minimum
A closed can must hold \(250\pi\text{ cm}^3\). Find the minimum-area design, then compare it with a feasible design having radius \(10\text{ cm}\).
For the minimum-area design,
Its height is \(h=2r=10\text{ cm}\). Its volume checks: \(\pi(5\text{ cm})^2(10\text{ cm})=250\pi\text{ cm}^3\). Its end area is \(2\pi(5\text{ cm})^2=50\pi\text{ cm}^2\), and its side area is \(2\pi(5\text{ cm})(10\text{ cm})=100\pi\text{ cm}^2\). Thus its total area is \(150\pi\text{ cm}^2\). Also, \(2\pi(5\text{ cm})^3=250\pi\text{ cm}^3\), confirming the critical condition.
For the alternative radius \(r=10\text{ cm}\), the volume constraint requires
This alternative also has the required volume: \(\pi(10\text{ cm})^2(2.5\text{ cm})=250\pi\text{ cm}^3\). Its ends have area \(2\pi(10\text{ cm})^2=200\pi\text{ cm}^2\), and its side has area \(2\pi(10\text{ cm})(2.5\text{ cm})=50\pi\text{ cm}^2\). Its total is \(250\pi\text{ cm}^2\), which exceeds the minimum \(150\pi\text{ cm}^2\). The comparison is consistent with the global result: the alternative radius differs from the unique minimizing radius.
A Common Error: Losing the Volume Factor
When substituting the height into the area formula, keep the entire fixed volume in the numerator. The side area becomes \(2\pi r(V/(\pi r^2))=2V/r\), not a term with an arbitrary fraction of \(V\).
For example, with \(V=54\pi\text{ cm}^3\), the correct reduced objective is
Using \(18\pi\text{ cm}^3/r\) instead would give \(A(3\text{ cm})=18\pi\text{ cm}^2+6\pi\text{ cm}^2=24\pi\text{ cm}^2\), not the actual \(54\pi\text{ cm}^2\). The volume check exposes the error: at \(r=3\text{ cm}\) and \(h=6\text{ cm}\), the side area is \(2\pi rh=36\pi\text{ cm}^2\), and the reduced side-area term must give that same value.
Interpretation and Scope
The result \(h=2r\) balances the contributions of the curved side and circular ends at the optimal dimensions. At that design, the side contributes \(2\pi r(2r)=4\pi r^2\), while the two ends contribute \(2\pi r^2\). The side therefore uses twice the surface area of the two ends together. This is a consequence of the optimization, not a claim that the three-dimensional parts use equal amounts of material.
The conclusion depends on the model: the cylinder is closed, its volume is fixed, and material use is represented by total surface area. Different construction requirements can change the objective. Within the stated model, however, the derivative sign proves that the dimensions found are the unique global minimum.
Check Your Understanding
Use the fixed-volume constraint and the surface-area model to answer the following.
- For a closed cylinder of radius \(r\) and height \(h\), write its volume and total surface-area formulas.
- For fixed volume \(V>0\), solve the volume constraint for \(h\) and obtain the one-variable area function.
- What critical-point equation results from differentiating the reduced area function?
- Why does the derivative change from negative to positive at the candidate radius?
- For a fixed volume, what relation connects the height and radius of the minimum-area can?
