1st Putnam 1938 Problem A2 A solid has a cylindrical middle with a conical cap at each end. The height of each cap equals the length of the middle. For a given surface area, what shape maximizes the volume?
Let the radius be R and the height H. The area is 2πRH + 2πR√(R 2 +H 2 ). The volume is 5/3 πR 2 H. The area is fixed, so for some fixed k, we have R ( H + √(R 2 +H 2 ) ) = k. This gives H = (k 2 - R 4 )/(2kR). We must now choose R to maximise f(R) = R 2 H = R (k 2 - R 4 )/2k. Evidently the allowed range for R is from R = 0 up to √k (corresponding to H = 0). But f(0) = 0 and f(√k) = 0, so the maximum is at some interior point of the interval. Differentiating, we find it is at R max = (k 2 /5) 1/4 . In terms of the area A, we have A = 2πk, so R max = (A/(π 2√5) ) 1/2 . Many thanks to Philip Ascher for correcting an error in the last line! 1st Putnam 1938 © John Scholes jscholes@kalva.demon.co.uk 5 Mar 2002 Last corrected/updated 22 Feb 03