A cylinder shaped can needs to be constructed to hold 400 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 0.07 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 Volume of a cylinder: V = Tr²h Area of the sides: A = 2rrh Area of the top/bottom: A To minimize the cost of the can: The radius of the can should be: The minimum cost should be: cents The height of the can should be:

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Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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### Optimization Problem: Designing a Cost-Efficient Soup Can

A cylinder-shaped can needs to be constructed to hold 400 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 needs to be thicker and costs 0.07 cents per square centimeter. Find the dimensions for the can that will minimize production cost.

#### Helpful Information:
- \( h \) : height of the can
- \( r \) : radius of the can

**Volume of a cylinder:**
\[ V = \pi r^2 h \]

**Area of the sides:**
\[ A = 2 \pi r h \]

**Area of the top/bottom:**
\[ A = \pi r^2 \]

### To minimize the cost of the can:
**The radius of the can should be:**
\[ \text{(Fill in the value)} \]

**The minimum cost should be:**
\[ \text{(Fill in the value)} \text{ cents} \]

**The height of the can should be:**
\[ \text{(Fill in the value)} \]

### Explanation of Diagrams and Graphs (if any):

_(Note: Since this problem does not include any diagrams or graphs, this section can be omitted for now. If there were diagrams or graphs, such as a cost versus dimensions graph, you would explain the axes, important points, and any relevant trends here.)_

### How to Approach the Problem:
1. **Start with the Volume Equation:**
   - The volume equation \( V = \pi r^2 h \) should be set equal to 400 cubic centimeters: 
   \[ 400 = \pi r^2 h \]
   
2. **Express height \( h \) in terms of \( r \):**
   - Rearrange the equation \( 400 = \pi r^2 h \) to solve for \( h \):
   \[ h = \frac{400}{\pi r^2} \]

3. **Calculate the Cost Equation:**
   - The cost includes the material for both the sides and the top/bottom of the can. Using the areas given:
     - Cost of the sides: \( 0.04 \times 2\pi rh \)
     - Cost of the top and bottom: \( 0.07 \times 2
Transcribed Image Text:### Optimization Problem: Designing a Cost-Efficient Soup Can A cylinder-shaped can needs to be constructed to hold 400 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 needs to be thicker and costs 0.07 cents per square centimeter. Find the dimensions for the can that will minimize production cost. #### Helpful Information: - \( h \) : height of the can - \( r \) : radius of the can **Volume of a cylinder:** \[ V = \pi r^2 h \] **Area of the sides:** \[ A = 2 \pi r h \] **Area of the top/bottom:** \[ A = \pi r^2 \] ### To minimize the cost of the can: **The radius of the can should be:** \[ \text{(Fill in the value)} \] **The minimum cost should be:** \[ \text{(Fill in the value)} \text{ cents} \] **The height of the can should be:** \[ \text{(Fill in the value)} \] ### Explanation of Diagrams and Graphs (if any): _(Note: Since this problem does not include any diagrams or graphs, this section can be omitted for now. If there were diagrams or graphs, such as a cost versus dimensions graph, you would explain the axes, important points, and any relevant trends here.)_ ### How to Approach the Problem: 1. **Start with the Volume Equation:** - The volume equation \( V = \pi r^2 h \) should be set equal to 400 cubic centimeters: \[ 400 = \pi r^2 h \] 2. **Express height \( h \) in terms of \( r \):** - Rearrange the equation \( 400 = \pi r^2 h \) to solve for \( h \): \[ h = \frac{400}{\pi r^2} \] 3. **Calculate the Cost Equation:** - The cost includes the material for both the sides and the top/bottom of the can. Using the areas given: - Cost of the sides: \( 0.04 \times 2\pi rh \) - Cost of the top and bottom: \( 0.07 \times 2
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