Question
there are two lemonade stands that sell lemonade called stand A and stand B. The total cost for lemonade at Stand A is $15 (C) plus $2 per ice (I). The total cost for lemonade at Stand B is $17 (C) plus $4 per ice (I). Which Lemonade stand is cheaper. Lets use C for the total cost and I for the cost per ice.
Systems of linear equations can be used to solve many, but not all, real-world problems. When you chose the problem for your journal entry, how did you know that this problem could be solved using a system of linear equations? What specific characteristics does your problem have that made it a good choice? List at least three reasons why the problem you described in your journal prompt can be solved using a system of linear equations.
Systems of linear equations can be used to solve many, but not all, real-world problems. When you chose the problem for your journal entry, how did you know that this problem could be solved using a system of linear equations? What specific characteristics does your problem have that made it a good choice? List at least three reasons why the problem you described in your journal prompt can be solved using a system of linear equations.
Answers
GPT-4o mini
To determine which lemonade stand is cheaper, let's analyze the total cost equations for both stands:
**For Stand A:**
- Total Cost (C_A) = $15 + $2 * I
- Where \( I \) is the number of ice units purchased.
**For Stand B:**
- Total Cost (C_B) = $17 + $4 * I
### Comparing Costs:
To find out which stand is cheaper, we can set up an inequality or calculate the total costs based on the number of ice units. However, it may be helpful to find the point at which both stands have the same total cost:
Set \( C_A = C_B \):
\[ 15 + 2I = 17 + 4I \]
### Solving the Equation:
1. Subtract \( 2I \) from both sides:
\[ 15 = 17 + 2I \]
2. Subtract 17 from both sides:
\[ -2 = 2I \]
3. Divide by 2:
\[ I = -1 \]
Since we cannot have a negative number of ice units, this indicates that Stand A is cheaper when purchasing any ice (since the cost of ice at Stand A increases at a slower rate).
### Cost Comparison for Various Ice Units:
- If \( I = 0 \):
- Stand A: \( C_A = 15 \)
- Stand B: \( C_B = 17 \)
- Stand A is cheaper.
- If \( I = 1 \):
- Stand A: \( C_A = 15 + 2(1) = 17 \)
- Stand B: \( C_B = 17 + 4(1) = 21 \)
- Stand A is cheaper.
- If \( I = 2 \):
- Stand A: \( C_A = 15 + 2(2) = 19 \)
- Stand B: \( C_B = 17 + 4(2) = 25 \)
- Stand A is cheaper.
- Continuing this way, you find that Stand A remains the cheaper option.
### Reasons Why This Problem Can be Solved Using a System of Linear Equations:
1. **Linear Relationships:** Both total costs for Stand A and Stand B can be expressed as linear equations in terms of the variable \( I \). These linear equations represent the relationship between the number of ice units and the total cost.
2. **Two Variables:** The problem involves two independent lemonade stands that can be represented by two separate equations, allowing for the comparison of their costs under the same variable \( I \). It provides a clear way to analyze and compare both scenarios simultaneously.
3. **Point of Intersection:** The ability to find a point where both equations yield the same total cost (even if it leads to an impractical solution like negative ice) suggests that a system of equations is appropriate. It shows there's a point at which the costs equalize and directs attention to the comparison of both stands across various conditions.
This problem exemplifies how systems of linear equations can effectively represent and solve real-world scenarios involving comparative costs, making it suitable for analysis using mathematical methods.
**For Stand A:**
- Total Cost (C_A) = $15 + $2 * I
- Where \( I \) is the number of ice units purchased.
**For Stand B:**
- Total Cost (C_B) = $17 + $4 * I
### Comparing Costs:
To find out which stand is cheaper, we can set up an inequality or calculate the total costs based on the number of ice units. However, it may be helpful to find the point at which both stands have the same total cost:
Set \( C_A = C_B \):
\[ 15 + 2I = 17 + 4I \]
### Solving the Equation:
1. Subtract \( 2I \) from both sides:
\[ 15 = 17 + 2I \]
2. Subtract 17 from both sides:
\[ -2 = 2I \]
3. Divide by 2:
\[ I = -1 \]
Since we cannot have a negative number of ice units, this indicates that Stand A is cheaper when purchasing any ice (since the cost of ice at Stand A increases at a slower rate).
### Cost Comparison for Various Ice Units:
- If \( I = 0 \):
- Stand A: \( C_A = 15 \)
- Stand B: \( C_B = 17 \)
- Stand A is cheaper.
- If \( I = 1 \):
- Stand A: \( C_A = 15 + 2(1) = 17 \)
- Stand B: \( C_B = 17 + 4(1) = 21 \)
- Stand A is cheaper.
- If \( I = 2 \):
- Stand A: \( C_A = 15 + 2(2) = 19 \)
- Stand B: \( C_B = 17 + 4(2) = 25 \)
- Stand A is cheaper.
- Continuing this way, you find that Stand A remains the cheaper option.
### Reasons Why This Problem Can be Solved Using a System of Linear Equations:
1. **Linear Relationships:** Both total costs for Stand A and Stand B can be expressed as linear equations in terms of the variable \( I \). These linear equations represent the relationship between the number of ice units and the total cost.
2. **Two Variables:** The problem involves two independent lemonade stands that can be represented by two separate equations, allowing for the comparison of their costs under the same variable \( I \). It provides a clear way to analyze and compare both scenarios simultaneously.
3. **Point of Intersection:** The ability to find a point where both equations yield the same total cost (even if it leads to an impractical solution like negative ice) suggests that a system of equations is appropriate. It shows there's a point at which the costs equalize and directs attention to the comparison of both stands across various conditions.
This problem exemplifies how systems of linear equations can effectively represent and solve real-world scenarios involving comparative costs, making it suitable for analysis using mathematical methods.