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Minimizing cost-weighted surface material for a fixed-volume can

A manufacturer makes a cylindrical soup can that must hold a fixed volume of 355355 mL (i.e., 355 cm3355 \text{ cm}^3). Because the top and bottom require thicker, reinforced metal, they cost twice as much per unit area as the side. Let rr be the radius and hh be the height of the can, both in centimeters.

The cost (in arbitrary units) of the metal used to make the can is C(r,h)=2πr2k+2πrh,C(r,h) = 2\pi r^2 k + 2\pi r h, where k=2k=2 accounts for the top and bottom costing twice as much per square centimeter as the lateral side. The volume constraint is πr2h=355.\pi r^2 h = 355. Use the constraint to eliminate hh, and find the radius r>0r>0 that minimizes the total cost C(r)C(r).

Hint

Solve the volume constraint for hh in terms of rr, substitute into C(r,h)C(r,h) to get a function of rr alone, then find where C′(r)=0C'(r)=0 and confirm it's a minimum.

Step-by-step solution

From the constraint πr2h=355\pi r^2 h = 355, solve for hh: h=355πr2h = \dfrac{355}{\pi r^2}. Substituting into C(r,h)=2πr2(2)+2πrhC(r,h) = 2\pi r^2 (2) + 2\pi r h gives C(r)=4πr2+2πr⋅355πr2=4πr2+710r,C(r) = 4\pi r^2 + 2\pi r \cdot \frac{355}{\pi r^2} = 4\pi r^2 + \frac{710}{r}, valid for r>0r>0.

Differentiate: C′(r)=8πr−710r2C'(r) = 8\pi r - \dfrac{710}{r^2}. Setting C′(r)=0C'(r)=0 gives 8πr3=7108\pi r^3 = 710, so r3=7108π=3554πr^3 = \dfrac{710}{8\pi} = \dfrac{355}{4\pi}, hence r=(3554π)1/3r = \left(\dfrac{355}{4\pi}\right)^{1/3}.

Since C′(r)<0C'(r) < 0 for small rr and C′(r)>0C'(r) > 0 for large rr (or checking C′′(r)=8π+1420r3>0C''(r) = 8\pi + \frac{1420}{r^3} > 0), this critical point is a minimum. Numerically, r≈(3554π)1/3≈2.94r \approx \left(\frac{355}{4\pi}\right)^{1/3} \approx 2.94 cm minimizes the cost.

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