A cylinder shaped can needs to be constructed to hold 350 cubic centimeters of soup. The material for the sides of the can costs 0.04 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.08 cents per square centimeter. Find the dimensions for the can that will minimize production cost.

Helpful information:
*h* : height of can, *r* : radius of can

to minimize the cost of the can: 

radius of the can: ____ 
height of the can: ____ 
minimum cost: _____

To minimize the production cost of the cylindrical can, we need to find the optimal dimensions (radius \( r \) and height \( h \)) while considering the given cost constraints and the volume of the can.

First, we have the volume constraint:
\[ V = \pi r^2 h = 350 \]

We need to express the height \( h \) in terms of the radius \( r \):
\[ h = \frac{350}{\pi r^2} \]

Next, let's calculate the cost of the materials used. The cost consists of the cost of the sides, the top, and the bottom of the can. 

1. The surface area of the sides is \( 2\pi rh \).
2. The surface area of the top and bottom is \( 2\pi r^2 \).

Given the costs:
- The cost of the sides is \( 0.04 \) cents per square centimeter.
- The cost of the top and bottom is \( 0.08 \) cents per square centimeter.

So, the total cost \( C \) can be expressed as:
\[ C = 0.04 \times 2\pi rh + 0.08 \times 2\pi r^2 \]
\[ C = 0.08\pi rh + 0.16\pi r^2 \]

Substitute \( h \) from the volume constraint into the cost function:
\[ C = 0.08\pi r \left( \frac{350}{\pi r^2} \right) + 0.16\pi r^2 \]
\[ C = 0.08 \times \frac{350}{r} + 0.16\pi r^2 \]
\[ C = \frac{28}{r} + 0.16\pi r^2 \]

To find the value of \( r \) that minimizes the cost, we take the derivative of \( C \) with respect to \( r \) and set it to zero:
\[ \frac{dC}{dr} = \frac{d}{dr} \left( \frac{28}{r} + 0.16\pi r^2 \right) \]
\[ \frac{dC}{dr} = -\frac{28}{r^2} + 0.32\pi r \]

Set the derivative equal to zero:
\[ -\frac{28}{r^2} + 0.32\pi r = 0 \]
\[ 0.32\pi r = \frac{28}{r^2} \]
\[ 0.32\pi r^3 = 28 \]
\[ r^3 = \frac{28}{0.32\pi} \]
\[ r^3 = \frac{28}{0.32 \times 3.14159} \]
\[ r^3 \approx \frac{28}{1.00532} \]
\[ r^3 \approx 27.849 \]
\[ r \approx \sqrt[3]{27.849} \]
\[ r \approx 3.037 \]

Now, we calculate the height \( h \) using the volume constraint:
\[ h = \frac{350}{\pi r^2} \]
\[ h = \frac{350}{\pi (3.037)^2} \]
\[ h = \frac{350}{3.14159 \times 9.224} \]
\[ h \approx \frac{350}{28.996} \]
\[ h \approx 12.073 \]

Finally, we calculate the minimum cost:
\[ C = \frac{28}{3.037} + 0.16\pi (3.037)^2 \]
\[ C \approx 9.22 + 0.16\pi \times 9.224 \]
\[ C \approx 9.22 + 0.16 \times 28.996 \]
\[ C \approx 9.22 + 4.639 \]
\[ C \approx 13.859 \] cents

So, the optimal dimensions and the minimum cost are:
- Radius of the can: \( \approx 3.037 \) cm
- Height of the can: \( \approx 12.073 \) cm
- Minimum cost: \( \approx 13.859 \) cents

Would you like more details or have any questions?

Here are five questions you might want to ask next:
1. Can you explain how the cost function was derived from the surface areas?
2. Why did we take the derivative of the cost function and set it to zero?
3. How do you interpret the solution in terms of cost efficiency?
4. Can we verify the solution by checking the second derivative?
5. What would happen if the costs of the materials changed?

Tip: Always check if your solution meets the given constraints, such as volume or cost, when solving optimization problems.

Discover how to minimize production costs for a cylindrical soup can holding 350 cubic centimeters. Calculate optimal dimensions using geometry and calculus, considering material costs per square centimeter. Suitable for advanced high school students.

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