# math

posted by
**ashish**
.

A right cicular cylindrical can is to be constructed to have a volume of 57.749 cubic inches (one quart). The sides of the can are to be formed by rolling and welding a strip of metal, which may be purchased in rolls with width equal to the desired height of the can. The material for the sides costs 20 cents per square foot. The welding cost is 1.1 cents per inch. Top and bottom of the can are circles cut from hexagons to minimize the waste. the width of the hexagon is the diameter of the can plus 0.4 inches. the extra inches is crimpled over the sides to form the seal. Crimping costs are 1.6 cents per inch and the material for the ends sells for 30 cents per square foot. The metal for the ends of the cans may be purchased in a rolls which allows for exactly 4 hexagons and exactly 3 hexagons in alternate strips. Find the dimensions of the most economical can which can be constructed to meet these specifications.

Write equations for the amount of both rolls of material needed to produce seven cans, as a function of the diameter of the can. The height can be expressed in terms of the required volume, V, and the diameter, D.

V = (1/4) pi D^2 h

h = 4 V/[pi D^2]

Then compute the total cost of the material and processing for seven cans, as a function of diameter only, with V as a constant.

Differentiate Cost vs(D) and set the derivative = 0. Then solve for the optimum diameter.

The reason you should do this for seven cans is that you get seven lids at a time, with minimum waste. Remember than seven cans require a total of 14 tops and bottoms