A cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0.02 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.06 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 πr² h = Area of the sides: A = 2πrh Area of the top/bottom: A πr ² = To minimize the cost of the can: Radius of the can: Height of the can: Minimum cost: cents
A cylinder shaped can needs to be constructed to hold 200 cubic centimeters of soup. The material for the sides of the can costs 0.02 cents per square centimeter. The material for the top and bottom of the can need to be thicker, and costs 0.06 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 πr² h = Area of the sides: A = 2πrh Area of the top/bottom: A πr ² = To minimize the cost of the can: Radius of the can: Height of the can: Minimum cost: cents
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter9: Systems Of Equations And Inequalities
Section9.4: Linear Programming
Problem 23E
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![### Optimizing the Design of a Cylindrical Can
In this example, we will determine the optimal dimensions for a cylindrical can to hold 200 cubic centimeters of soup, with the aim of minimizing the production cost.
**Problem Statement:**
We need to construct a cylindrical can with the following characteristics:
- Volume of the can: 200 cubic centimeters.
- The material for the sides costs 0.02 cents per square centimeter.
- The material for the top and bottom costs 0.06 cents per square centimeter.
**Objective:**
Find the dimensions of the can (radius and height) that will minimize the production cost.
#### Helpful Information:
- \( h \): height of the can
- \( r \): radius of the can
##### Formulas Involved:
1. **Volume of a Cylinder:**
\[ V = \pi r^2 h \]
2. **Area of the Sides:**
\[ A_{\text{sides}} = 2\pi rh \]
3. **Area of the Top/Bottom:**
\[ A_{\text{top/bottom}} = \pi r^2 \]
##### Steps to Minimize the Cost:
1. **Equate the Volume:**
\[ 200 = \pi r^2 h \]
2. **Calculate the Total Cost:**
- Cost for the sides:
\[ \text{Cost}_{\text{sides}} = 2\pi rh \times 0.02 \]
- Cost for the top and bottom:
\[ \text{Cost}_{\text{top/bottom}} = 2 \times \pi r^2 \times 0.06 \]
3. **Combine the Costs:**
The total production cost is:
\[ \text{Total Cost} = (2\pi rh \times 0.02) + (2\pi r^2 \times 0.06) \]
4. **Substitute \( h \) from the volume formula into the cost formula:**
\[ h = \frac{200}{\pi r^2} \]
5. **Derive the Cost Function:**
Substitute \( h \) in the total cost formula and solve for \( r \).
6. **Differentiate and Find Critical Points:**
Differentiate the cost function with respect to \( r \) and find the critical points to minimize the cost.
##### Dimensions](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe345ae0b-fd83-4d2e-a467-23dbb110ff23%2F649a2c38-0ba5-4a8a-bc8d-8b5b9ae7ef48%2Fyi97csj_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Optimizing the Design of a Cylindrical Can
In this example, we will determine the optimal dimensions for a cylindrical can to hold 200 cubic centimeters of soup, with the aim of minimizing the production cost.
**Problem Statement:**
We need to construct a cylindrical can with the following characteristics:
- Volume of the can: 200 cubic centimeters.
- The material for the sides costs 0.02 cents per square centimeter.
- The material for the top and bottom costs 0.06 cents per square centimeter.
**Objective:**
Find the dimensions of the can (radius and height) that will minimize the production cost.
#### Helpful Information:
- \( h \): height of the can
- \( r \): radius of the can
##### Formulas Involved:
1. **Volume of a Cylinder:**
\[ V = \pi r^2 h \]
2. **Area of the Sides:**
\[ A_{\text{sides}} = 2\pi rh \]
3. **Area of the Top/Bottom:**
\[ A_{\text{top/bottom}} = \pi r^2 \]
##### Steps to Minimize the Cost:
1. **Equate the Volume:**
\[ 200 = \pi r^2 h \]
2. **Calculate the Total Cost:**
- Cost for the sides:
\[ \text{Cost}_{\text{sides}} = 2\pi rh \times 0.02 \]
- Cost for the top and bottom:
\[ \text{Cost}_{\text{top/bottom}} = 2 \times \pi r^2 \times 0.06 \]
3. **Combine the Costs:**
The total production cost is:
\[ \text{Total Cost} = (2\pi rh \times 0.02) + (2\pi r^2 \times 0.06) \]
4. **Substitute \( h \) from the volume formula into the cost formula:**
\[ h = \frac{200}{\pi r^2} \]
5. **Derive the Cost Function:**
Substitute \( h \) in the total cost formula and solve for \( r \).
6. **Differentiate and Find Critical Points:**
Differentiate the cost function with respect to \( r \) and find the critical points to minimize the cost.
##### Dimensions
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