# 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

## To find the dimensions of the most economical can, we will first need to write equations for the amount of material needed to produce seven cans, as a function of the diameter of the can.

Let's use the following variables:
- V: Required volume of one can (57.749 cubic inches)
- D: Diameter of the can
- h: Height of the can

Firstly, we can express the height, h, in terms of the required volume, V, and the diameter, D:
h = (4V) / (πD^2)

Secondly, we need to consider the material for the sides of the can. The sides are formed by rolling and welding a strip of metal, which has a width equal to the desired height of the can. The cost of this material is 20 cents per square foot.

We can calculate the amount of material required for the sides as follows:
- Length of one side strip = circumference of the can = πD
- Width of one side strip = height of the can = h
- Area of one side strip = Length * Width = πD * h

Since we need to produce seven cans, the total amount of material needed for the sides is 7 times the area of one strip:
Material for sides = 7 * (πD * h)

Next, let's consider the material for the top and bottom of the can. The top and bottom are circles cut from hexagons, and the width of the hexagon is the diameter of the can plus 0.4 inches. The extra inches are crimped 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.

We can calculate the amount of material required for the ends as follows:
- Length of the hexagon side = circumference of the can = πD
- Width of the hexagon = diameter of the can plus 0.4 inches = D + 0.4
- Area of one hexagon = Length * Width = πD * (D + 0.4)
- Since we get 4 hexagons for every 7 cans, the total amount of material needed for the ends is 4/7 times the area of one hexagon:
Material for ends = (4/7) * (πD * (D + 0.4))

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

Total cost = (Material cost for sides + Material cost for ends) + (Welding cost + Crimping cost)

Since the costs are given in different units (cents per square foot and cents per inch), we also need to convert the dimensions to the appropriate units before calculating the costs.

Finally, we can differentiate the total cost with respect to D (the diameter) and set the derivative equal to zero. By solving this equation, we can find the optimum diameter that minimizes the total cost and therefore gives us the most economical can.