I am stumped. Find the ratio of height to radius that minimizes the surface of a right cylinder with a volume of Vm3(cubed)

Find an expression for height in terms of volume.
Take the formula for the surface area of the cylinder and substitute in the height expression from above.
Take the first derivative of this equation.
Set the equation equal to 0 to find the minimum.
Solve for V
Substitute in the formula for volume.
Solve for h.
You should have a formula in the form of h = xr
That's your ratio!

And you're done
Let me know the answer you get!

OK I got -r = h in the end don't know if that's right, but thanks cause you helped a lot.

I'm hoping that it's obvious to you that your answer is wrong, you cannot have a negative dimension :)

You should get h = 2r

Do you want to post your working and I'll go through it for you?

In any case you have had a good stab at it and hopefully your teacher can help you now :)

Volume = V = pi * r^2 * h, so h = V/(pi * r^2).
Ssurface area = S = 2 * pi * r^2 + 2 * pi * r * h = 2 * pi * r^2 + 2 * pi * r * V / (pi * r^2)
= 2 * pi * r^2 + 2 * V / r.
So dV/dr = 4 * pi * r - 2 * V / r^2 = (4 * pi * r^3 - 2 * V) / r = 0 iff r = (V / (2 * pi))^1/3.
Then h = V / (pi * r^2) = V/(pi * (V / (2 * pi))^2/3 = 2^2/3 * V^1/3 / pi^1/3
= (4 * V / pi)1/3
Hence h / r = (4 * V / pi)^1/3 / (V / (2 * pi))^1/3 = 4^1/3

Elisha, I would love to check your answer and work out where you went wrong, but it's such a darn mess that I can't make head nor tail of it. In the middle of it, where you calculate dV/dr, do you mean dS/dr, since that is what we're trying to minimise (I notice you have the dV/dr in terms of V, so I assume that's what you meant.)?
If you could tidy it up a little then I would be happy to help you work out where you went wrong.

Again, I would like to point you to This Announcement that states the site's policy on helping people who are asking homework questions.

Hope you will reply :)

Elisha, I would suggest if you're going to post to any extent, learn a little LaTex.

I am trying to sound critical, but it's just a struggle to read something like that.

... [1]

... [2]

Solve [2] for h:



Sub into [1]:



Differentiate:



Set to 0 and solve for r:

... [3]

Then ... [4]

[3]/[4]=1/2

In other words, as Capuchin said, the radius is 1/2 the height. That is, the surface area is minimized when the diameter is equal to the height.

**Please disregard those 'squigglies'. They came about when I tried to enlarge the font.