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.

Question

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.

Answer 1

To find the dimensions of the most economical can, we need to minimize the cost of materials and welding.

Let's start by understanding the different components of the can:

1. Sides of the can: The sides are formed by rolling and welding a strip of metal. The cost of the material for the sides is 20 cents per square foot, and the welding cost is 1.1 cents per inch.

2. Top and bottom of the can: The top and bottom are circles cut from hexagons to minimize waste. 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. The crimping cost is 1.6 cents per inch, and the material for the ends sells for 30 cents per square foot.

Now, let's proceed with finding the dimensions of the can step by step:

1. Start by defining the variables:
- Let "r" be the radius of the can.
- Let "h" be the height of the can.

2. Volume of the can:
The volume of the right circular cylindrical can is given as 57.749 cubic inches (one quart). The volume formula is: V = πr^2h.
Substituting the given volume, we have: 57.749 = πr^2h.

3. Cost of the sides:
The cost of the sides is given by the area of the metal strip multiplied by the cost per square foot. The area of the strip is the circumference multiplied by the height of the can. Therefore, the cost of the sides is: sides_cost = 2πrh * 12 * 20 cents.

4. Cost of crimping:
The cost of crimping is the perimeter of the top/bottom circle multiplied by the crimping cost per inch. The perimeter of the circle is 2πr. Therefore, the cost of crimping is: crimping_cost = 2πr * 1.6 cents.

5. Cost of the top and bottom:
The cost of the top and bottom is the area of the hexagon multiplied by the cost per square foot. The area of the hexagon is given by the formula: (3√3/2) * (r + 0.4)^2 square inches. Therefore, the cost of the top and bottom is: top_bottom_cost = (3√3/2) * (r + 0.4)^2 * 4 * 30 cents.

6. Total cost:
The total cost is the sum of the cost of the sides, crimping cost, and the cost of the top and bottom. Therefore, the total cost is: total_cost = sides_cost + crimping_cost + top_bottom_cost.

7. With the constraints in mind, we need to minimize the total cost by finding the values of "r" and "h" that satisfy the given conditions.

Applying these steps and solving the equations will give us the dimensions of the most economical can which can be constructed to meet the specified specifications.