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March 26, 2015

March 26, 2015

Posted by **Jus** on Saturday, January 11, 2014 at 9:59am.

- math -
**Reiny**, Saturday, January 11, 2014 at 12:02pmIn 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

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