Why a Fixed Volume Creates an Optimization Problem
A cylindrical container can be made tall and narrow or short and wide while holding the same amount. These designs do not use the same amount of material: changing the radius and height changes both the area of the circular ends and the area of the curved side.
We will find the dimensions that use the least surface area when the cylinder is closed—it has a top and a bottom—and its volume is fixed. The model assumes the material has uniform thickness and that seams, overlap, and waste are not included in the area calculation.
Build the Model
The two circular ends each have area \(\pi r^2\), so together they contribute \(2\pi r^2\). Unrolling the curved side gives a rectangle with height \(h\) and width equal to the circle's circumference, \(2\pi r\). Its area is therefore \(2\pi rh\). Thus the closed cylinder's surface area is \(S=2\pi r^2+2\pi rh\), while its volume is \(V=\pi r^2h\).
Because the volume is fixed, the radius and height cannot vary independently. Solve the volume constraint for \(h\):
Substituting this into surface area makes area a function of the single variable \(r\):
The domain is all positive radii. For every such radius, the constraint gives exactly one positive height, so every \(r>0\) corresponds to a feasible cylinder. The optimization problem is to minimize \(S(r)\) over that entire domain.
Worked Example: A Cylinder Holding \(54\pi\) Cubic Centimeters
Worked Example: Find the Minimum-Material Dimensions
A closed cylinder must have volume \(54\pi\text{ cm}^3\). Find the dimensions that minimize its surface area.
Here \(V=54\pi\). Eliminate height using the volume constraint:
Substitute into the surface-area formula:
Differentiate and set the derivative equal to zero to find an interior candidate:
The corresponding height is \(h=54/3^2=54/9=6\text{ cm}\). The area is \(2\pi(3^2)+2\pi(3)(6)=18\pi+36\pi=54\pi\text{ cm}^2\). Check the volume: \(\pi(3^2)(6)=54\pi\text{ cm}^3\), as required. The height is \(6=2(3)\), twice the radius. The minimum claim follows from the general feasibility and minimum argument below.
Derive the Optimal Proportions
For any prescribed \(V>0\), differentiate the one-variable objective. Since \(r>0\), the derivative is defined throughout the feasible domain:
A critical point must satisfy \(S'(r)=0\). The denominator \(r^2\) is positive, so this is equivalent to \(4\pi r^3-2V=0\). Solving gives one positive candidate radius, and the volume constraint then determines its height.
To see why the height relationship follows, use \(V=2\pi r^3\), which holds at the critical radius, in the constraint formula for height: \(h=V/(\pi r^2)=(2\pi r^3)/(\pi r^2)=2r\). This is the relationship between the dimensions; it does not say that the height is twice the radius for every cylinder with the same volume.
Why the Critical Point Is a Feasible Absolute Minimum
A critical point alone does not establish an absolute minimum. The candidate must be in the feasible domain, and the behavior of the objective on that domain must be checked. Here the critical radius \((V/(2\pi))^{1/3}\) is positive because \(V>0\), so it is feasible. We can also show directly that the objective decreases before that radius and increases after it.
\(S'(r)=0\) exactly when \(r^3=V/(2\pi)\). Since \(r^3\) increases strictly for \(r>0\), there is exactly one such radius.
For \(r\) smaller than the critical radius, \(4\pi r^3-2V<0\), so \(S'(r)<0\). For \(r\) larger, \(4\pi r^3-2V>0\), so \(S'(r)>0\).
The surface area decreases up to the critical radius and increases after it. Thus this feasible point gives the unique minimum over all \(r>0\), not just a local minimum.
The ends of the feasible domain support the same conclusion. As \(r\) approaches zero from above, the term \(2V/r\) grows without bound. As \(r\) grows without bound, \(2\pi r^2\) grows without bound. Thus a cylinder cannot achieve the minimum by becoming arbitrarily narrow and tall or arbitrarily wide and short.
At the optimum, the lateral area is \(2\pi rh=2\pi r(2r)=4\pi r^2\), and the two ends together have area \(2\pi r^2\). So the total minimum area can also be written as \(6\pi r^2\), with \(r=(V/(2\pi))^{1/3}\).
Worked Example: Compare Two Feasible Designs
Worked Example: Check Which Design Uses Less Material
Two closed cylinders each hold \(16\pi\text{ cm}^3\). One has radius \(1\text{ cm}\); the other has radius \(4\text{ cm}\). Compare their surface areas with the minimum.
For the optimal design, the radius satisfies \(r^3=V/(2\pi)=16\pi/(2\pi)=8\), so \(r=2\text{ cm}\). Its height is \(h=16\pi/(\pi\cdot 2^2)=16/4=4\text{ cm}\), and its area is \(2\pi(2^2)+2\pi(2)(4)=8\pi+16\pi=24\pi\text{ cm}^2\). Its volume checks: \(\pi(2^2)(4)=16\pi\text{ cm}^3\).
For the cylinder with \(r=1\text{ cm}\), the height required by the constraint is \(h=16\pi/(\pi\cdot1^2)=16\text{ cm}\). Its area is \(2\pi(1^2)+2\pi(1)(16)=2\pi+32\pi=34\pi\text{ cm}^2\). Its volume is \(\pi(1^2)(16)=16\pi\text{ cm}^3\).
For the cylinder with \(r=4\text{ cm}\), the height is \(h=16\pi/(\pi\cdot4^2)=16/16=1\text{ cm}\). Its area is \(2\pi(4^2)+2\pi(4)(1)=32\pi+8\pi=40\pi\text{ cm}^2\). Its volume is \(\pi(4^2)(1)=16\pi\text{ cm}^3\).
The three feasible areas are \(24\pi\), \(34\pi\), and \(40\pi\text{ cm}^2\), respectively. Both alternatives use more material than the optimal design. The derivative argument establishes that no other positive radius can give a smaller area.
Worked Example: A Volume That Does Not Give an Integer Radius
Worked Example: Keep the Exact Result Before Rounding
Find the dimensions of the minimum-area closed cylinder with volume \(100\text{ cm}^3\). Give numerical dimensions and area to three decimal places.
The optimal radius satisfies \(r^3=100/(2\pi)=50/\pi\), so the exact radius is \(r=(50/\pi)^{1/3}\text{ cm}\). The optimal height is \(h=2(50/\pi)^{1/3}\text{ cm}\). Numerically, these are \(r\approx2.515\text{ cm}\) and \(h\approx5.031\text{ cm}\), rounded to three decimal places.
The exact minimum area is \(6\pi r^2=6\pi(50/\pi)^{2/3}\text{ cm}^2\), which is approximately \(119.265\text{ cm}^2\) to three decimal places.
Check the volume using the exact expressions, before rounding: \(\pi r^2h=\pi(50/\pi)^{2/3}\cdot2(50/\pi)^{1/3}=2\pi(50/\pi)=100\text{ cm}^3\). The exact expressions satisfy the constraint; the rounded dimensions need not reproduce it exactly because rounding changes their values.
A Common Modeling Mistake
A frequent error is to use the surface-area formula for an open-top cylinder when the container is closed. For example, one might write \(S_{\text{wrong}}=\pi r^2+2\pi rh\), counting only one circular end. After substituting \(h=V/(\pi r^2)\), this becomes \(S_{\text{wrong}}(r)=\pi r^2+2V/r\). Differentiating gives \(S'_{\text{wrong}}(r)=2\pi r-2V/r^2\), so the resulting critical-point equation is \(r^3=V/\pi\), which leads to \(h=r\).
That calculation is consistent with the incorrect model, but the model describes a cylinder with one end, not a closed cylinder. For a closed cylinder, the correct area is \(2\pi r^2+2\pi rh\). Its derivative gives \(r^3=V/(2\pi)\), and the corresponding height is \(h=2r\). Before differentiating, check that the objective function includes every surface specified in the problem.
Why the Proportion Matters
The result is useful because it gives a simple design rule independent of the particular positive volume: among closed cylinders holding a fixed amount, the minimum-area design has height equal to its diameter. The scale changes with volume—the optimal radius is \((V/(2\pi))^{1/3}\)—but the proportion does not.
The conclusion depends on the stated model. If the top is removed, if the material cost differs between surfaces, or if there are additional constraints on dimensions, the objective or feasible set changes and the optimal proportion need not be the same. For the closed, uniform-material cylinder considered here, eliminating height, differentiating on the feasible domain, and checking the derivative's sign establish the minimum.
Check Your Understanding
Use the closed-cylinder model and keep the feasibility condition \(r>0\) in mind.
- Write the surface-area formula for a closed cylinder and identify the contribution of each term.
- For fixed positive volume \(V\), eliminate height and write the surface area as a function of radius.
- What equation determines the critical radius, and what is the resulting relationship between height and radius?
- Why does the critical point give an absolute minimum over all feasible radii?
- What incorrect height-to-radius relationship results from omitting one circular end?
