# Thread: Applied Project The shape of a can

1. ## Applied Project The shape of a can

The most of the expense is incurred in joining the sides to the rims of the can. If we cut the discs from hexagons then the total cost is proportional to
4√3 r2 + 2pi*r*h + k(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= ((pih)/r)^1/3 * (2pi-h/r)/(pih/r-4*3^(1/3))

2. ## Shape of a can

Need help to complete this optimization problem.The most of the expense is incurred in joining the sides to the rims of the can. If we cut the discs from hexagons then the total cost is proportional to
4√3 r2 + 2pi*r*h + k(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= ((pih)/r)^1/3 * (2pi-h/r)/(pih/r-4*3^(1/3))
thank u for ur help in advance

3. ## Re: Applied Project The shape of a can

Can I just say how confusing this part of the problem is. So 4root3 times radius cubed and 2 times pi times radius times hight describe the surface area of the rectangular cutout and the two hexagons without the two circles, and I guess the length of the sum of the 2 circumferances and the hight are proportional to the surface area when multiplied by the constant K in this case. So am I right in reading the problem that the cost is proportional to the surface area? I still want to find the minimal surface area... lol has anybody else come across this problem in James Steewarts 3rd edition of single variable calculus? It is the applied project at the end of section 4.6. ...