Find the size of a can (\(R\) and \(h\)) that contains the maximum possible volume, given that its surface area is \(0.0300 \mathrm m^2\).
This problem can be solved using Lagrange multipliers. The volume of the can is
\( V = \pi R^2h \)
And the area gives us the imposed condition
\( A = 2\pi Rh + 2\pi R^2 = 0.0300 \mathrm m^2 \)
This can also be written as
\( \phi = 2\pi Rh + 2\pi R^2 - 0.0300 \mathrm m^2 = 0 \)
The maximum should happen when
\( \dfrac{\partial V}{\partial R} = \lambda \dfrac{\partial \phi}{\partial R} \)
and
\( \dfrac{\partial V}{\partial h} = \lambda \dfrac{\partial \phi}{\partial h} \)
Then the equations are:
\( 2\pi RH = \lambda(2\pi h + 4\pi R) \)
and
\( \pi R^2 = \lambda (2 \pi R) \)
Dividing one equation by the other (to eliminate lambda)
\( \dfrac {2\pi Rh} {\pi R^2} = \dfrac {2\pi h + 4\pi R}{2 \pi R} \)
\( \rightarrow 2 \dfrac hR = \dfrac {h+2R} {R} \)
\( \rightarrow h = 2R \)
Therefore, to get the maximum volume the height of the can has to be equal to
the diameter (how often do you see that in the supermarket?)
In our problem, this implies that
\( A =2 \pi Rh +2\pi R^2 = 6 \pi R^2 = 0.0300 \mathrm m^2 \rightarrow R = \sqrt{\dfrac{0.0300 \mathrm m}{6\pi}} = 0.04 \mathrm m \)
And \( h = 0.08 \mathrm m \)