Help setting up Lagrangian for fixed-volume cylinder surface minimization

[Abigail_T]

I'm working through a challenging calculus problem set for my engineering class, and I'm completely stuck on an optimization problem involving minimizing the surface area of a cylindrical can with a fixed volume. I've set up my primary equation and constraint, and I think I need to use Lagrange multipliers, but I keep getting tangled in the algebra when solving the system of partial derivatives. Could someone walk me through the steps of setting up the Lagrangian function for this specific scenario and the process of solving for the critical points? I understand the theory, but the application is tripping me up.

[VioletR]

Here's the setup and solution using Lagrange multipliers for a closed cylinder (radius r, height h). Objective: minimize S = 2π r^2 + 2π r h with the constraint V = π r^2 h (fixed). The Lagrangian is L(r,h,λ) = 2π r^2 + 2π r h + λ(π r^2 h - V).

Compute the partial derivatives:
∂L/∂r = 4π r + 2π h + λ(2π r h) = 0
∂L/∂h = 2π r + λ(π r^2) = 0
∂L/∂λ = π r^2 h - V = 0

From ∂L/∂h: 2π r + λ π r^2 = 0 ⇒ λ = −2/r (since r > 0).
Plug into ∂L/∂r: 4π r + 2π h + (−2/r)(2π r h) = 4π r + 2π h − 4π h = 0 ⇒ 4π r − 2π h = 0 ⇒ h = 2r.
Use the constraint: π r^2 h = V ⇒ π r^2 (2r) = V ⇒ 2π r^3 = V ⇒ r = (V/(2π))^{1/3}.
Then h = 2r = (4V/π)^{1/3}.

Conclusion: the minimum occurs at h = 2r, with r = (V/(2π))^{1/3} and h = (4V/π)^{1/3}. (If you’re wary of the second-derivative test, note this is the unique interior critical point on the constraint and S → ∞ as r → 0+ or ∞, so it’s the minimum on the feasible set.)

[Paisley_L]

Another way to see it (substitution): from V = π r^2 h, solve for h = V/(π r^2) and substitute into S to get S® = 2π r^2 + 2V/r. Differentiating and setting dS/dr = 0 gives 4π r^3 = 2V, hence r^3 = V/(2π) and r = (V/(2π))^{1/3}; then h = V/(π r^2) = (4V/π)^{1/3}, same result as above.

[Justin.M]

Open-top cylinder variant: if the top is not present, S = π r^2 + 2π r h with V = π r^2 h. The Lagrange setup yields 2π r + λ π r^2 = 0 and 2π r + 2π h + λ(2π r h) = 0, which leads to h = r and r = (V/π)^{1/3}. So the no-top case has a different optimal proportion (height equals radius) while the closed case has height equal to diameter.

[LoganSP]

Numerical check (example): take V = 1. Then r = (1/(2π))^{1/3} ≈ 0.542, h = 2r ≈ 1.084. The minimal surface area is S_min = 2π r^2 + 2π r h ≈ 6π r^2 ≈ 5.52. This matches the relationship h = 2r nicely.