00:01
For this problem, we are asked to find the dimensions of the closed right circular cylindrical can of smallest surface area, whose volume is 16 pi centimeters cubed.
00:11
So we are trying to minimize the surface area, which will be equal to 2 times pi r squared from the top and bottom, plus pi, or excuse me, plus 2 pi r h.
00:24
Subject to the constraint that, well, our volume equals 16 pi.
00:28
So that would mean that pi times r squared times h must be equal to, or actually we can write this as the constraint, g, equals pi r squared h minus 16 pi, which then in turn must be equal to zero.
00:43
So we set this up as a lagrange multiplier problem with gradient of s equals lambda times the gradient of g.
00:50
Taking the partial derivative first with respect to r gives us that 4 pi r plus 2 pi h must be equal to 2 pi r or excuse me 2 pi to lambda pi r h then taking the partial derivative with respect to h gives us that 2 pi r must be equal to lambda times pi r squared.
01:26
Dividing both, or actually, i won't get ahead.
01:29
In this case, we can safely divide both sides by r because it would be nonsensical to be talking about something with zero radius...