1. ## Lagrange Multipliers

Find the dimensions of the closed right circular cylindrical can of smallest surface area whose volume is $16pi\;cm^3$.
I managed to find r = 2cm, h = 4cm which gives the surface area of $24pi \,cm^2$ using the method $\nabla f=\lambda \nabla g \; and \; g(x,y,z)=0$.
My question is that, how come the final result of r and h is min and not max. What if the question ask for the largest surface area in this case?

2. You needn't Lagrange multipliers:

$\pi r^2h=16\Leftrightarrow h=16/\pi r^2$

now, substitute $h$ in $S$.

Regards.

Fernando Revilla

3. Thanks but the chapter is Lagrange multipliers so... and what's that S you're talking about?

4. Hello, tintin2006!

$\text{Find the dimensions of the closed right circular cylindrical can}$
$\text{of smallest surface area whose volume is }16\pi\text{ cm}^3$.

$\text{My question: how come the final result of }r\text{ and }h\text{ is min and not max?}$
$\text{What if the question asked for the }largest\text{ surface area?}$

Consider a lump of stuff about the size of a Campbell's soup can.
. . And let's say it has a volume of $16\pi$ cm $^2.$
We can measure the surface area of that cylinder.

Now take a rolling pin and flatten the stuff like a pizza.

We will still have a cylinder.
. . It will have a large circular top and bottom . . . and a very small height.
You can "see" that it has a much larger surface area.

In theory, we can continue to flatten it and make a larger and larger pizza.
It may be only one molecule thick, but its top and bottom may cover an acre.

Flatten it even more until its thickness is that of an atom.
By then the pizza dough may cover, say, Canada.

My point is that there is no maximum surface area for the cylinder.

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# Minimum surface area with fixed volumeFind the dimen-sions of the closed right circular cylindrical can of smallest sur-face area whose volume is 16pi.

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