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Calculus/Min. and Max. problem
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Question
Please help on the question,
What is the maximum volume of a cup of a given surface area S, in the form of a right circular cylinder with no top?
S = πr² + 2πrh where r is the radius and h is the height.
From here, it can be said that V = πr²h.
Solving the first equation for h gives h = (S - πr²)/(2πr).
Putting into V gives V = πr²(S - πr²)/(2πr).
The πr/πr cancels, leaving V = r(S - πr²)/2.
This is the same as V = rS/2 - πr^3/2.
Knowing this, it can be said that dV/dr = S/2 - 3πr²/3.
Setting this to 0 gives 0 = S/2 - 3πr²/2, so S/2 = 3πr²/2.
From here, we can solve for r = √(S/3π).
Since we had S = πr² + 2πrh, we can say that h = (S - πr²)/(2πr).
Puting in what r is gives h = (S - π(S/3π))/(2π√(2S/(3π))).
h = ( S - S/3)(2π√(2S)/√(3π)), which after a few steps gives
h = √(6πS)/12πS.
Putting this into V = πr²h gives
V = π(S/(3π))(√(6πS)/12πS)
= √(6πS)/(36π).
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