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Calculus and Beyond Homework Help
Maximizing the volume of a cylinder
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[QUOTE="BvU, post: 6332183, member: 499340"] Yes it's needed ... I'd like to see what you use as Lagrangian and what you use as ##q_k## and ##g_k## [edit] You seem to think that ##{\mathcal L} = f = R^2 ## but that is not correct Furthermore, the number of constraints is not equal to the number of variables, so ##g_k## and ##\lambda_k## looks strange: The sum is not over ##k## but over the constraints :##\displaystyle{ \sum_i \lambda_i \frac{\partial g_i}{\partial q_k}}## The ##\lambda_i## are numbers, not functions. - - - - - - - - - - - - - - - [edit2] looked at your profile and want to step back somewhat: ##f## is the function you want to maximize, so here ## f(r,h) = V = 2\pi r^2 h ##, a function of 2 variables. At an extreme you expect ##{\partial f\over \partial h}= {\partial f\over \partial r} = 0##. Two equations in two unknowns that are not independent. Instead of eliminating one of the variables, you use the method of [URL='https://en.wikipedia.org/wiki/Lagrange_multiplier']Lagrange multipliers[/URL] and solve ##{\partial {\mathcal L} \over \partial h}= {\partial {\mathcal L} \over \partial r} = 0## with ##{\mathcal L} = f - \lambda g##. ##g(r,h) = 0## is the constraint ##S = 2\pi r^2 + 2\pi rh \quad \Rightarrow \quad g(r,h) = 2\pi r^2 + 2\pi rh - S##. Now we have three equations (the third one is ##g=0##) with three unknowns (##r,h,\lambda##) and we can treat ##r## and ##h## as if they were independent. -- [/QUOTE]
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Maximizing the volume of a cylinder
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