Posted by Jus on Saturday, January 11, 2014 at 9:59am.
In my diagram , I arbitrarily let the radius of the cylinder be 1 unit
let the radius of the cylinder be r , and let the height of the cylinder be 2h
That way I can say:
r^2 + h^2 = 1
r^2 = 1 - h^2 or r = (1- h^2)^(1/2)
Surface area (SA) = 2πr^2 + 2πrh
= 2π(1-h^2) + 2π(1-h^2)^(1/2) h
= 2π(1-h^2) + 2π (h^2 - h^4)^(1/2)
d(SA)/dh = 2π [-2h + (1/2)(h^2 - h^4)((-1/2) (2h - 4h^3) ]
= 0 for a max of SA
(1/2)(2h-4h^3)/√(h^2 - h^4) = 2h
(1/2)(2h)(1 - 2h^2) / (h√(1-h^2) ) = 2h
(1 - 2h^2)/√(1 - h^2) = 2h
1 - 2h^2 = 2h√(1-h^2)
square both sides
1 - 4h^2 + 4h^4 = 4h^2 - 4h^4
8h^4 - 8h^2 + 1 = 0
solving this I got
h^2 = (2 ± √2)/4
case1: h^2 = (2+√2)/4
h = ..9238796 , r = ..38268.. , h/r = 2.41421.. or 1 + √2
case2: h^2 = (2-√2)/4
h = .38268... , r = .9238.. , h/r = .4142 .. or -1 + √2
Looking at Wolfram
http://www.wolframalpha.com/input/?i=maximize+2π%281-h%5E2%29+%2B+2π+%28h%5E2+-+h%5E4%29%5E%281%2F2%29
I will take case 2 as my answer.
Nasty, nasty question