# 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.

Please solve this. I tried almost 1 hour, but I don't get it.

## To find the dimensions of the most economical can, we need to minimize the cost of both the sides and the ends of the can.

Let's break down the problem step by step:

1. Start by determining the dimensions of the can. Let's assume the height of the can is 'h' inches, and the diameter is 'd' inches.

2. Calculate the volume of the can using the given information that it should be 57.749 cubic inches:
Volume of a right circular cylinder = π * (d/2)^2 * h
Therefore, π * (d/2)^2 * h = 57.749

3. Calculate the surface area of the side of the can:
Surface area of a right circular cylinder = 2 * π * (d/2) * h

4. Determine the dimensions of the hexagon from which the top and bottom (ends) of the can will be cut. The width of the hexagon is given as the diameter of the can plus 0.4 inches.

5. Calculate the area of a hexagon using the formula:
Area of a hexagon = (3√3 / 2) * s^2, where 's' is the side length of the hexagon.
Since the hexagon is inscribed in a circle, the side length 's' is equal to the radius of the circle. Therefore, s = d/2.
Substituting s = d/2 in the area formula, we get: Area of a hexagon = (3√3 / 2) * (d/2)^2

6. Determine the dimensions of the hexagon such that there is the least waste. The extra inches are crimpled over the sides to form the seal.

7. Calculate the cost of material for the sides of the can. The material costs 20 cents per square foot, and we need to convert the side's surface area from square inches to square feet.

8. Calculate the cost of welding the sides of the can. The welding cost is given as 1.1 cents per inch.

9. Calculate the cost of material for the ends of the can. The material costs 30 cents per square foot. Note that the metal for the ends of the cans may be purchased in rolls allowing for exactly 4 hexagons and exactly 3 hexagons in alternate strips.

10. Calculate the cost of crimping. The cost of crimping is given as 1.6 cents per inch.

11. Add up the costs of material, welding, and crimping to obtain the total cost of constructing the can.

12. Use mathematical optimization techniques, such as calculus, to minimize the total cost function by finding the values of 'd' and 'h' that make the cost as low as possible. Set up a function for the total cost in terms of 'd' and 'h' and find its minimum by taking partial derivatives.

13. Solve for the values of 'd' and 'h' that minimize the total cost.

By following these steps and performing the necessary calculations, you can determine the dimensions of the most economical can.