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Unread 13-03-2005, 21:33
sanddrag sanddrag is offline
On to my 16th year in FRC
FRC #0696 (Circuit Breakers)
Team Role: Teacher
 
Join Date: Jul 2002
Rookie Year: 2002
Location: Glendale, CA
Posts: 8,507
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Re: Proving smallest surface area of cylinder is when h=2r

Quote:
Originally Posted by Kevin Sevcik
sanddrag: V is a constant in this case since you said yourself for a given volume treat it as another number. Thus:
A' = 4*pi*r - 2*V/r^2
I can follow you here. The key IS to think of V as something like some random number (has a derivative of 0). Then when I set this equal to zero and solve I get 2V=4*pi*r^3. Then I substitue in pi*r^2*h for V and voila! it reduces down to 2h=r.



Thanks so much everyone!
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Teacher/Engineer/Machinist - Team 696 Circuit Breakers, 2011 - Present
Mentor/Engineer/Machinist, Team 968 RAWC, 2007-2010
Technical Mentor, Team 696 Circuit Breakers, 2005-2007
Student Mechanical Leader and Driver, Team 696 Circuit Breakers, 2002-2004
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