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Minimizing surface area of a cylinder with fixed volume

A cylindrical soda can must hold a fixed volume V=355 cm3V = 355 \text{ cm}^3 of liquid. The can consists of a circular top, a circular bottom, and a lateral (side) surface, and the manufacturer wants to use as little aluminum as possible.

Let the can have radius rr and height hh (both in cm). The volume constraint is πr2h=355\pi r^2 h = 355, and the total surface area (top, bottom, and side) is S=2πr2+2πrhS = 2\pi r^2 + 2\pi r h.

Find the radius rr (in cm) that minimizes the total surface area SS, and find the minimum surface area itself.

Hint

Solve the volume constraint for hh in terms of rr, substitute into the surface area formula to get SS as a function of rr alone, then find where S′(r)=0S'(r) = 0. Check that this critical point gives a minimum, for instance using the sign of S′′(r)S''(r) or the behavior of S′(r)S'(r) as r→0+r \to 0^+ and r→∞r \to \infty.

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 S=2πr2+2πrhS = 2\pi r^2 + 2\pi r h gives S(r)=2πr2+710rS(r) = 2\pi r^2 + \dfrac{710}{r}, defined for r>0r > 0. Differentiating, S′(r)=4πr−710r2S'(r) = 4\pi r - \dfrac{710}{r^2}.

Setting S′(r)=0S'(r) = 0 gives 4πr=710r24\pi r = \dfrac{710}{r^2}, so r3=7104π=177.5πr^3 = \dfrac{710}{4\pi} = \dfrac{177.5}{\pi}, giving r=(177.5π)1/3≈3.837r = \left(\dfrac{177.5}{\pi}\right)^{1/3} \approx 3.837 cm. Since S′′(r)=4π+1420r3>0S''(r) = 4\pi + \dfrac{1420}{r^3} > 0 for all r>0r>0, this critical point is a minimum. Plugging r≈3.837r \approx 3.837 back into S(r)=2πr2+710rS(r) = 2\pi r^2 + \dfrac{710}{r} gives the minimum surface area S≈277.5S \approx 277.5 cm2^2.

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