I know this is a homework problem but please keep reading. the question is: "Find the ratio of the radius R to height H of a right-circular cylinder of fixed volume V that minimizes the surface area A" now this is easy, and i can do this by just mimimising the surface area with respect to the ratio R/H, PLEASE DONT POST THIS SOLUTION I CAN DO THIS. Now the problem i have is that i cant figure how to approach this from a calculus of variations point of view(which is what the question asks) I cant even come up with an integral that that gives me the entire surface area. i tried it not using the entire surface area but only using the curved surface area and saying my unknown function f(volume) = R/H but when i substituted in the euler-langrange equation i just got that r->inf so i obviously need an integral that gives the entire surface area. any suggestions appreciated

what? im taking pre-cal and ive done somethin like this before, but i have no idea what u want! try not to use so many big words.

Folks,

Please note that vivi2k is following the correct forum procedure for a homework question here. Although they did not show much work, they described their approach.

More importantly, vivi2k is not asking for an answer, merely suggestions .

I''ll allow the thread to remain open.

Please do not provide answers, only suggestions.

Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

Its been many years since I studied the Calculus of Variations, and without a review I don''t have any real suggestions. But I did find a link that may be helpful:

planetmath.org/encyclopedia/CalculusOfVariations.html



Graham Rhodes
Senior Scientist
Applied Research Associates, Inc.

Thanks, I have checked out the link, but i can do that part(the other questions i had to do were similar to that). my problem is forming the actual integral im supposed to minimize. i''ll go and see my lecturer next week.

As far as I know there isn''t a integral, but rather the sum of two integrals for the surface area of a cylindar. There is a integral for the volume. A cylindar is a surface of revolution, specifically a line swept around an axis.

LilBudyWizer "As far as I know there isn't a integral, but rather the sum of two integrals for the surface area of a cylindar" - I know, thats my problem but im thinking because the volume is constant i can get H in terms of R or vice versa so i can make the integration variables the same, and if i can fiddle the limits i can add the integrands together therefore only one integral.

AP - Im sorry if something i wrote upset you but i dont think i deserved that. i dont normally post work related questions on this forum but i was just reading it earlier and thought to myself i wonder if anyone had any ideas. and in the future i know not to ask this sort of thing again.

[edited by grhodes_at_work: I removed the AP post. It was inappropriate an unacceptable.]

[edited by - grhodes_at_work on January 24, 2003 12:42:26 PM]

Hi

I would do it the following way:

given: V = constneed:  a = r/hsome cylinder formulas:(1) V = pi*r²*h(2) A = 2* 2*pi*r² + 2*r*husing(2) yields :(3) h = V/(pi*r²)inserting (3) in (2)A = 4*pi*r² + 2*V/(pi*r)--> get minimum using differentiating --> rminusing(3)(4) hmin = V/(pi*rmin²)finally:a = rmin / hmin 

Bye
ScottManDeath

I don''t know what "calculus of variations" is exactly, but I know how to set up a surface integral. A surface integral is a double integral- ie you integrate over two dimensions.

For example, if I wanted to compute the surface area of the square with length 4 and width 5, the integral would look like: (where S stands for that big integral sign)

4 5
S S (1) dx dy
0 0

So the integral goes from 0 to 5 over x, and 0 to 4 over y, and when you work it out the answer is obviously 20.

(The 1 in the middle is the function of integration- since we are just doing area we leave it at 1, but you can do interesting stuff with a more complex function)

So your goal will be to set up 3 integrals (one for the side, and twice the circle at the top/bottom), and add them together to get the area.

ScottManDeath - Cheers, but i have done it a similar way to this already. the point is i have to do it using calculus of variations.

AP - Calculus of variations requires that you set up a SINGLE integral and fiddle with the integrand. i had thought of doing a surface integral but i cant think of a way to parametize the surface such that i get both the curved surface area and the flat surface area in the same integral.