1. ## Cylindrical Can Optimization

This is my first post and involves optimization of the most economic shape of a can. The problem verbatim is "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 as in task #4, then the total cost is proportional to

4*3^(1/3)r^2+2pirh+k(4pir+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= ((pih)/r)^1/3 * (2pi-h/r)/(pih/r-4*3^(1/3))

I have no idea where to begin with this one. Any help is much appreciated.

2. Originally Posted by heebee
This is my first post and involves optimization of the most economic shape of a can. The problem verbatim is "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 as in task #4, then the total cost is proportional to

$4*3^{1/3}r^2+2\pi rh+k(4\pi r+h)$

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:

$V^{1/3}/k= ((\pi h)/r)^{1/3} * (2\pi-h/r)/(\pi h/r-4*3^{1/3})$

I have no idea where to begin with this one. Any help is much appreciated.
Use the fact that $V=\pi r^2h$ to replace h by $V/(\pi r^2)$ in the expression for the total cost. Then the cost is (proportional to) $f(r) = 4*3^{1/3}r^2+2V/r +k(4\pi r+ (V/(\pi r^2)))$. To minimise that, differentiate with respect to r, put the derivative equal to 0, and you will get an equation for k. You should find that this equation can be written as $\frac rk = \frac{2\pi - (h/r)}{V/r^3 -4*3^{1/3}}$.

From $V=\pi r^2h$ it follows that $r^3 = \frac{Vr}{\pi h}$ and so $r = \Bigl(\frac{Vr}{\pi h}\Bigr)^{1/3}$. Substitute that expression for r into the left side of the previous formula, and you should get the right result.