The cheapest can
Tonight's idea
A can must hold a fixed volume \(V\). Which shape uses least metal? With radius \(r\) and height \(h\), the volume is \(V = \pi r^2 h\) and the surface area is \(A = 2\pi r^2 + 2\pi r h\): two discs plus the side, a rectangle \(2\pi r\) by \(h\) when unrolled.
Two variables, one constraint. Use the constraint to eliminate one: \(h = V/(\pi r^2)\), so
\[A(r) = 2\pi r^2 + \frac{2V}{r}.\]
The first term grows with \(r\) (metal on the ends), the second shrinks (metal on the side). Between them the slope is zero:
\[A'(r) = 4\pi r - \frac{2V}{r^2} = 0 \quad\Longrightarrow\quad r^3 = \frac{V}{2\pi}.\]
Put this back: \(h = V/(\pi r^2) = 2\pi r^3/(\pi r^2) = 2r\). The best can is exactly as tall as it is wide.
Is it a minimum? \(A''(r) = 4\pi + 4V/r^3 > 0\) for every \(r > 0\), so the curve bends upwards and the flat point is its lowest. The check is part of the answer: a zero slope could equally mark a maximum.
Write the quantity in one variable using the constraint. Set the derivative to zero. Confirm with \(A''\), and look at the ends of the allowed range.
Watch one: a 330 ml can
\(V = 330\) cm³. Then \(r^3 = 330/(2\pi) \approx 52.5\), so \(r \approx 3.74\) cm and \(h = 2r \approx 7.49\) cm. The metal used is
\[A = 2\pi(3.74)^2 + \frac{660}{3.74} \approx 264 \text{ cm}^2.\]
Cans in a shop are taller and narrower than this. Our model charges every square centimetre the same, but the ends are stamped from thicker metal than the side. Exercise 4 puts that in.
Your turn
- Two positive numbers multiply to 100. Write their sum as a function of one of them, minimise it, and show your answer is a minimum rather than a maximum. (Compare AM–GM.)
- An open-topped box has a square base of side \(x\) and must hold 4 m³. Write the area of sheet metal as a function of \(x\), minimise it, and give the dimensions of the box and the metal used.
- A pot has a base but no lid, and must hold a volume \(V\). Show that the least metal is used when \(h = r\), half the proportion of the closed can. Explain in one sentence why dropping the lid makes the best pot wider.
- Back to the closed can, but now the two ends cost twice as much per cm² as the side. Minimise the cost, not the area. Show that the cheapest can has \(h = 4r\), twice as tall as it is wide, and say which term in the cost pushed the answer that way.
Stretch: the end of the range
A wire of length \(L\) is cut in two. One piece is bent into a circle, the other into a square. Where should you cut to make the total area as small as possible? Then: where should you cut to make it as large as possible, and why does the derivative miss that answer?
Answers
- \(S(x) = x + 100/x\), \(S'(x) = 1 - 100/x^2 = 0\) at \(x = 10\) (we need \(x > 0\)). \(S''(x) = 200/x^3 > 0\), so a minimum. Both numbers are 10, sum 20.
- Height \(h = 4/x^2\), so \(A = x^2 + 4xh = x^2 + 16/x\). \(A' = 2x - 16/x^2 = 0\) gives \(x^3 = 8\), \(x = 2\) m, \(h = 1\) m. \(A'' = 2 + 32/x^3 > 0\): a minimum. Metal used \(4 + 8 = 12\) m².
- \(A = \pi r^2 + 2V/r\), \(A' = 2\pi r - 2V/r^2 = 0\) gives \(r^3 = V/\pi\), and \(h = V/(\pi r^2) = \pi r^3/(\pi r^2) = r\). \(A'' = 2\pi + 4V/r^3 > 0\). With one disc instead of two, widening the pot costs half as much, so the balance tips towards a wider base.
- Cost \(C = 2(2\pi r^2) + 2\pi r h = 4\pi r^2 + 2V/r\). \(C' = 8\pi r - 2V/r^2 = 0\) gives \(r^3 = V/(4\pi)\), so \(h = V/(\pi r^2) = 4\pi r^3/(\pi r^2) = 4r\). The doubled ends term \(4\pi r^2\) punishes radius harder, so the optimum trades radius for height.
Stretch. Give the circle \(x\) of wire, so its radius is \(x/(2\pi)\) and its area \(x^2/(4\pi)\); the square has side \((L - x)/4\) and area \((L-x)^2/16\). \(A'(x) = x/(2\pi) - (L-x)/8 = 0\) gives \(x = \pi L/(4 + \pi) \approx 0.44L\) to the circle, with \(A'' = 1/(2\pi) + 1/8 > 0\): the minimum. The maximum is at an end of the range, \(x = L\): all of the wire in the circle, area \(L^2/(4\pi) \approx 0.080L^2\), beating the all-square \(L^2/16 = 0.0625L^2\). The derivative only finds flat points inside the range. Endpoints have to be checked by hand.