The typical soft drink can in the U.S. has a volume of 355 cm3. The two circular ends cost $0.0008 per cm2 each (because they are thicker), and the cost of the aluminum for the side of the can is $0.00026 per cm2. What dimensions will minimize the cost of a can? (Round your answers to two decimal places.) 
What is the minimum cost per can? (Round your answer to four decimal places.)

To minimize the cost of the can, we need to find the dimensions (radius and height) that minimize the total cost of the aluminum used for the can, subject to the constraint that the volume of the can is 355 cm³. The can is cylindrical, so the volume and surface area formulas will be key here.

### Given:
- Volume of the can, $ V = 355 \, \text{cm}^3 $
- Cost per unit area of the circular ends, $ C_{\text{ends}} = 0.0008 \, \text{per cm}^2 $
- Cost per unit area of the side, $ C_{\text{side}} = 0.00026 \, \text{per cm}^2 $

### Step 1: Volume of the Cylinder
The volume $ V $ of a cylinder is given by:
\[
V = \pi r^2 h
\]
where $ r $ is the radius and $ h $ is the height of the cylinder.

### Step 2: Surface Area of the Can
The surface area consists of two parts:
- The two circular ends: $ 2 \pi r^2 $
- The side of the can (lateral surface area): $ 2 \pi r h $

Thus, the total surface area is:
\[
A = 2 \pi r^2 + 2 \pi r h
\]

### Step 3: Total Cost of the Can
The total cost of the aluminum used in the can is the sum of the costs of the circular ends and the side:
\[
\text{Total cost} = 0.0008(2 \pi r^2) + 0.00026(2 \pi r h)
\]
\[
\text{Total cost} = 0.0016 \pi r^2 + 0.00052 \pi r h
\]

### Step 4: Solve for $ h $ in Terms of $ r $ Using the Volume Constraint
From the volume formula:
\[
h = \frac{355}{\pi r^2}
\]

### Step 5: Substitute $ h $ into the Cost Function
Substitute $ h = \frac{355}{\pi r^2} $ into the total cost function:
\[
\text{Total cost} = 0.0016 \pi r^2 + 0.00052 \pi r \left( \frac{355}{\pi r^2} \right)
\]
Simplify the expression:
\[
\text{Total cost} = 0.0016 \pi r^2 + \frac{0.00052 \times 355}{r}
\]
\[
\text{Total cost} = 0.0016 \pi r^2 + \frac{0.1846}{r}
\]

### Step 6: Minimize the Total Cost
To minimize the total cost, take the derivative of the cost function with respect to $ r $ and set it equal to zero:
\[
\frac{d}{dr} \left( 0.0016 \pi r^2 + \frac{0.1846}{r} \right) = 0
\]
The derivative is:
\[
0.0032 \pi r - \frac{0.1846}{r^2} = 0
\]
Multiply through by $ r^2 $ to simplify:
\[
0.0032 \pi r^3 = 0.1846
\]
Solve for $ r $:
\[
r^3 = \frac{0.1846}{0.0032 \pi}
\]
\[
r^3 \approx 18.36
\]
\[
r \approx 2.62 \, \text{cm}
\]

### Step 7: Solve for $ h $
Now, using $ r = 2.62 \, \text{cm} $ in the volume equation to solve for $ h $:
\[
h = \frac{355}{\pi (2.62)^2}
\]
\[
h \approx 16.45 \, \text{cm}
\]

### Step 8: Calculate the Minimum Cost
Substitute $ r = 2.62 \, \text{cm} $ and $ h = 16.45 \, \text{cm} $ back into the cost function:
\[
\text{Total cost} = 0.0016 \pi (2.62)^2 + \frac{0.1846}{2.62}
\]
\[
\text{Total cost} \approx 0.0342 + 0.0704 = 0.1046
\]
Thus, the minimum cost per can is approximately $ \boxed{0.1046} \, \text{dollars} $ or 10.46 cents.

### Final Answer:
- Radius: $ r \approx 2.62 \, \text{cm} $
- Height: $ h \approx 16.45 \, \text{cm} $
- Minimum cost per can: $ 0.1046 \, \text{dollars} $

Would you like more details or have any questions?

Here are 5 related questions you could explore:
1. How does changing the volume of the can affect its cost-minimizing dimensions?
2. What if the cost of the circular ends increased, how would that change the optimal dimensions?
3. How would you minimize the surface area instead of the cost for a fixed volume?
4. How does the shape of the can affect its structural strength and material use?
5. How does the cost function change if the side material has a different cost per cm²?

**Tip:** When solving optimization problems, always check the critical points by differentiating and setting the derivative equal to zero.

The typical soft drink can in the U.S. has a volume of 355 cm3. The two circular ends cost $0.0008 per cm2 each (because they are thicker), and the cost of the aluminum for the side of the can is $0.00026 per cm2. What dimensions will minimize the cost of a can? (Round your answers to two decimal places.) What is the minimum cost per can? (Round your answer to four decimal places.)

Discover how to minimize the cost of manufacturing a soft drink can with a volume of 355 cm³. Using calculus and geometry, this problem provides a detailed step-by-step solution to find the optimal radius, height, and minimum cost per can. Perfect for students in Grades 11-12 studying optimization problems.

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