Math · Calculus I · Worked example
Minimize the surface area of a can
A cylindrical can must hold 1000 cm³ (1 L). What radius and height use the least metal, counting the top and bottom?
Use the volume to remove h
The volume is fixed: πr²h = 1000. Solve for the height: h = 1000/(πr²).
Write the surface area in terms of r
The metal is the two circular ends plus the side: S = 2πr² + 2πrh. Substituting h gives a function of r alone, for r > 0.
Set the derivative to zero
S′(r) = 4πr − 2000/r². Setting it to zero gives 4πr³ = 2000, so r³ = 500/π and r = ∛(500/π) ≈ 5.42 cm.
Confirm a minimum
S″(r) = 4π + 4000/r³ is positive for every r > 0, so the graph is concave up and the critical point is a minimum. It is the only critical point, so it is the least surface area of all.
Find the height
Since πr³ = 500, h = 1000/(πr²) = 2r ≈ 10.84 cm: the height equals the diameter. The surface area is about 554 cm².
Result
r = ∛(500/π) ≈ 5.42 cm and h = 2r ≈ 10.84 cm, using about 554 cm² of metal.
Your turn
An open box is made from a 12 cm by 12 cm square of cardboard by cutting equal squares of side x from the corners and folding up the sides. Which x gives the largest volume?
Show the answer and explanation
x = 2 cm, for a volume of 128 cm³.
V(x) = x(12 − 2x)² for 0 < x < 6. V′(x) = (12 − 2x)(12 − 6x), which is zero at x = 2 (x = 6 leaves no box). V(2) = 2 · 8² = 128 cm³, while V is 0 at both ends.
Keep exploring
Open the graph of S(r): the minimum sits at r ≈ 5.42, and the area climbs steeply for thin, tall cans with small r.
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