Cylindrical Tanks With Unequal Material Costs

Why Different Material Costs Change the Design

A closed cylindrical tank uses material for its curved side and for its two circular ends. If the materials have different costs per unit area, minimizing the tank’s total surface area is not the same as minimizing its total cost. The least-cost design may be taller or shorter than the familiar design in which every part has the same unit cost.

Let \(r\) be the tank’s radius and \(h\) its height. We assume the tank is a true cylinder with \(r>0\) and \(h>0\), has a fixed positive capacity \(V\), and is made using uniform side and end materials with positive costs per unit area \(c_s\) and \(c_e\), respectively. The tank is closed, so both circular ends are included.

The objective is to minimize \(C\) among all positive \(r\) and \(h\) satisfying the capacity constraint. The material costs are positive constants; this model does not include additional costs such as seams, labor, or material waste.

Deriving the Optimal Shape

The fixed-capacity condition makes \(r\) and \(h\) dependent: choosing a radius determines the height required to hold volume \(V\). We can find the optimal shape using the Lagrange multiplier method for one equality constraint. At an interior constrained extremum, when the constraint gradient is nonzero, the objective gradient must be a scalar multiple of the constraint gradient.

Set \(g(r,h)=\pi r^2h\). Both partial derivatives of \(g\) are positive for \(r,h>0\), so its gradient is nonzero on the feasible set. The Lagrange equations, with variables ordered as \((r,h)\), are:

Thus, for some multiplier \(\lambda\), the equations are \(C_r=\lambda g_r\) and \(C_h=\lambda g_h\). The derivative \(4\pi c_er\) in \(C_r\) matters: the end-material cost changes when the radius changes, even if the height is held fixed.

Since \(r>0\), the second equation gives \(\lambda=2c_s/r\). Substituting this into the first gives the relationship between the optimal height and radius:

Now use the capacity constraint to determine the dimensions. Substitution of \(h=(2c_e/c_s)r\) into \(\pi r^2h=V\) gives the optimal radius; the constraint or the height-to-radius relationship then gives the optimal height.

Why This Candidate Is the Global Minimum

The multiplier calculation identifies an interior candidate, but by itself it does not establish that the candidate minimizes cost. To check the global behavior, use the capacity condition to express height in terms of radius:

Substituting this expression in the cost gives a one-variable function on the entire feasible set:

Its first and second derivatives are

Because \(c_s,c_e,V,r\) are positive, \(C''(r)>0\) for every \(r>0\). Thus \(C'\) is strictly increasing and can have at most one zero. It tends to negative infinity as \(r\) approaches zero from above, and to positive infinity as \(r\) increases without bound, so it has exactly one zero. Setting \(C'(r)=0\) gives \(r^3=c_sV/(2\pi c_e)\), the radius already obtained from the Lagrange equations.

Moreover, \(C(r)\) tends to infinity both as \(r\) approaches zero and as \(r\) increases without bound. Since the unique critical point is a minimum and the cost grows toward either end of the feasible radius range, it is the unique global minimum. This reduction also checks that no feasible radius has been excluded: every \(r>0\) determines exactly one feasible height \(V/(\pi r^2)\).

Worked Example: A Side Material That Costs Twice as Much

Worked Example: More Expensive End Material

Worked Example: Checking Nearby Feasible Designs

A Common Derivative Mistake

A frequent error is to differentiate only the side-material term with respect to \(r\). That term does depend on \(r\), but so does the end-material cost \(2\pi c_er^2\). For example, the calculation \(C_r=2\pi c_sh\) is wrong because it omits the derivative of the end cost.

Using the incorrect derivative in the first Lagrange equation, while keeping the correct second equation, would give \(\lambda=2c_s/r\) and then \(2\pi c_sh=4\pi c_sh\), forcing \(h=0\). That contradicts the positive capacity constraint. The contradiction is a warning that the derivative was incomplete, not a feature of the tank design. Differentiating every cost term before forming the multiplier equations prevents this error.

Interpreting the Result

The ratio \(h_*/r_*=2c_e/c_s\) describes how the relative prices affect the shape. If end material is more expensive relative to side material, the optimal tank is taller relative to its radius. If side material is more expensive, the optimal tank is shorter relative to its radius. These are relative comparisons: the absolute dimensions also depend on the required capacity.

When the two unit costs are equal, \(c_s=c_e\), the ratio becomes \(h_*/r_*=2\), or \(h_*=2r_*\). Unequal costs therefore adjust the familiar equal-cost shape by changing the balance between the side and end contributions. At the optimum, the cost of the side material is \(4\pi c_er_*^2\), while the cost of both ends is \(2\pi c_er_*^2\), so the side cost is twice the combined end cost.

These formulas are useful for preliminary design when capacity and material prices are known. They describe the optimum under the stated idealized model; real construction may impose additional requirements or costs that change the design problem.