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Proving smallest surface area of cylinder is when h=2r
I need to prove that surface area A of a cylinder (of volume V) is smallest when the height, h equals twice the radius, r.
We have two equations
1. V=(pi)(r^2)(h)
2. A=2(pi)(r^2) + 2(pi)(r)(h)
Here's what I think I need to do in order to prove the smallest A occurs when h=2r. I think I need to take equation 1 and solve for h which comes out to h=V/(pi)(r^2). Then I need to substitue that in for h in equation 2. This would make equation 2 become:
A=2(pi)(r^2) + 2(pi)(r)[V/(pi)(r^2)]
Now, I know this can be simplified, but I'm not sure if I'm doing it right. I think it would go to:
A=2(pi)(r^2) + 2V/r
Is this right? If so, what do I need to do now? I know I have to take the derivative to find the absolute minium. This is where I'm having trouble. Can someone please help? Thanks.
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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
Last edited by sanddrag : 13-03-2005 at 20:48.
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