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  "body": "\n## Problem\n\nLet $a \u003c b$ and $A, B \u003e 0$.\nAmong curves $y \\in C^1([a, b])$ with $y \u003e 0$, $y(a) = A$ and $y(b) = B$, find one whose [[surface of revolution|NE09EDK19V]] about the $x$-axis has the least [[surface area|0S29Y4ZW0Y]]\n$$S(y) = 2\\pi \\int_a^b y(x) \\sqrt{1 + y'(x)^2} \\, dx.$$\n\n## Remarks\n\nThe Lagrangian $y\\sqrt{1 + y'^2}$ does not depend on $x$, so the [[Beltrami identity|F7EKTEESQ5]] gives $y / \\sqrt{1 + y'^2} = c$ for a constant $c \u003e 0$.\nThe solutions are the catenaries $y(x) = c \\cosh\\big((x - d)/c\\big)$, and their surfaces of revolution are [[catenoids|6NN1HP54GB]].\nDepending on the positions of the two end points, there are two, one or no catenaries through them.\nWhen the end circles are far apart compared with their radii, no catenary joins the points and there is no minimiser among such smooth curves.\nThen the infimum of $S$ is $\\pi(A^2 + B^2)$, the area of the two end discs, approached by curves that run close to the axis; this limit is the Goldschmidt solution.\nFor some positions where catenaries do exist, the two discs still have smaller total area than every catenoid through the points, and again no smooth curve attains the infimum.\n\n## Explanations\n\n### L1 human\n\nDip two parallel rings into soapy water and pull them apart a little.\nThe soap film between them takes the shape of a catenoid, a surface of revolution whose area cannot be lowered by small changes.\nIf the rings are pulled too far apart, the film breaks and leaves a flat film on each ring.\n",
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