Thread: Can optimization problem

This is part of a larger can optimization problem. I've gotten the basic questions and all (like h/2r ratio), but I need help on this one. I'm not sure what to do with this question as it seems different than all the other optimization problems I've done thus far.

If we look more closely as some real cans, we see that the lid and the base are frequently formed from discs with radius larger than r that are bent over the ends of the can. If we allow for this, we would increase h/r. More significantly, in addition to the cost of the metal we need to incorporate the manufacturing of the can into the cost. Let’s assume that most of the expense is incurred in joining the sides to the rims of the cans. If we cut the discs from hexagons, then the total cost is proportional to:

4sqrt(3)r^2 + 2pi*r*h + k(4pi*r + h)

where k is the reciprocal of the length that can be joined for the cost of one unit area of metal. Show that this expression is minimized when:

V^(1/3) / k = ( ((pi*h)/r)^(1/3) ) * ( (2pi - h/r ) / ( (pi*h)/r - 4sqrt(3) )

Can anyone help me get started on this? I can't grasp my mind around this one.