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Liam and his friends are hosting a bake sale to raise funds for their school trip. They decided to sell cookies for $3 each and...Question
liam and his friends are hosting a bake sale to raise funds for their school trip. they decided to sell cookies for $3 each and muffins for $5 each their goal is to raise at least $300 but they want to sell no more than 80 items in total
what combination that Liam can sell to make at least $300 while keeping the total items sold to a maximum of 80
what combination that Liam can sell to make at least $300 while keeping the total items sold to a maximum of 80
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Answered by
GPT-4o mini
To determine the combinations of cookies and muffins Liam and his friends can sell to raise at least $300 while keeping the total items sold to a maximum of 80, we can set up the following equations based on the problem:
Let:
- \( x \) = number of cookies sold
- \( y \) = number of muffins sold
The constraints are:
1. \( 3x + 5y \geq 300 \) (to raise at least $300)
2. \( x + y \leq 80 \) (to sell no more than 80 items)
3. \( x \geq 0 \) (cannot sell negative items)
4. \( y \geq 0 \) (cannot sell negative items)
### Steps to find valid combinations:
1. From the second equation, we can express \( y \):
\[
y \leq 80 - x
\]
2. Substitute for \( y \) in the first equation:
\[
3x + 5(80 - x) \geq 300
\]
Simplifying:
\[
3x + 400 - 5x \geq 300
\]
\[
-2x + 400 \geq 300
\]
\[
-2x \geq -100
\]
\[
x \leq 50
\]
3. From \( y = 80 - x \), we substitute \( x \) to find \( y \):
- If \( x = 50 \):
\[
y = 80 - 50 = 30 \implies 3(50) + 5(30) = 150 + 150 = 300
\]
- If \( x = 0 \):
\[
y = 80 - 0 = 80 \implies 3(0) + 5(80) = 0 + 400 = 400
\]
4. Check combinations:
- Let's check integer values for \( x \) from 0 to 50:
- If \( x = 40 \):
\[
y = 80 - 40 = 40 \implies 3(40) + 5(40) = 120 + 200 = 320
\]
- If \( x = 30 \):
\[
y = 80 - 30 = 50 \implies 3(30) + 5(50) = 90 + 250 = 340
\]
- If \( x = 20 \):
\[
y = 80 - 20 = 60 \implies 3(20) + 5(60) = 60 + 300 = 360
\]
- If \( x = 10 \):
\[
y = 80 - 10 = 70 \implies 3(10) + 5(70) = 30 + 350 = 380
\]
- If \( x = 5 \):
\[
y = 80 - 5 = 75 \implies 3(5) + 5(75) = 15 + 375 = 390
\]
- If \( x = 0 \):
\[
y = 80 \implies 3(0) + 5(80) = 0 + 400 = 400
\]
### Conclusion:
Valid combinations of cookies and muffins that meet both criteria include:
1. \( (50, 30) \) - 50 cookies and 30 muffins
2. \( (40, 40) \) - 40 cookies and 40 muffins
3. \( (30, 50) \) - 30 cookies and 50 muffins
4. \( (20, 60) \) - 20 cookies and 60 muffins
5. \( (10, 70) \) - 10 cookies and 70 muffins
6. \( (5, 75) \) - 5 cookies and 75 muffins
7. \( (0, 80) \) - 0 cookies and 80 muffins
These combinations will allow them to meet their goal of raising at least $300 while selling no more than 80 items.
Let:
- \( x \) = number of cookies sold
- \( y \) = number of muffins sold
The constraints are:
1. \( 3x + 5y \geq 300 \) (to raise at least $300)
2. \( x + y \leq 80 \) (to sell no more than 80 items)
3. \( x \geq 0 \) (cannot sell negative items)
4. \( y \geq 0 \) (cannot sell negative items)
### Steps to find valid combinations:
1. From the second equation, we can express \( y \):
\[
y \leq 80 - x
\]
2. Substitute for \( y \) in the first equation:
\[
3x + 5(80 - x) \geq 300
\]
Simplifying:
\[
3x + 400 - 5x \geq 300
\]
\[
-2x + 400 \geq 300
\]
\[
-2x \geq -100
\]
\[
x \leq 50
\]
3. From \( y = 80 - x \), we substitute \( x \) to find \( y \):
- If \( x = 50 \):
\[
y = 80 - 50 = 30 \implies 3(50) + 5(30) = 150 + 150 = 300
\]
- If \( x = 0 \):
\[
y = 80 - 0 = 80 \implies 3(0) + 5(80) = 0 + 400 = 400
\]
4. Check combinations:
- Let's check integer values for \( x \) from 0 to 50:
- If \( x = 40 \):
\[
y = 80 - 40 = 40 \implies 3(40) + 5(40) = 120 + 200 = 320
\]
- If \( x = 30 \):
\[
y = 80 - 30 = 50 \implies 3(30) + 5(50) = 90 + 250 = 340
\]
- If \( x = 20 \):
\[
y = 80 - 20 = 60 \implies 3(20) + 5(60) = 60 + 300 = 360
\]
- If \( x = 10 \):
\[
y = 80 - 10 = 70 \implies 3(10) + 5(70) = 30 + 350 = 380
\]
- If \( x = 5 \):
\[
y = 80 - 5 = 75 \implies 3(5) + 5(75) = 15 + 375 = 390
\]
- If \( x = 0 \):
\[
y = 80 \implies 3(0) + 5(80) = 0 + 400 = 400
\]
### Conclusion:
Valid combinations of cookies and muffins that meet both criteria include:
1. \( (50, 30) \) - 50 cookies and 30 muffins
2. \( (40, 40) \) - 40 cookies and 40 muffins
3. \( (30, 50) \) - 30 cookies and 50 muffins
4. \( (20, 60) \) - 20 cookies and 60 muffins
5. \( (10, 70) \) - 10 cookies and 70 muffins
6. \( (5, 75) \) - 5 cookies and 75 muffins
7. \( (0, 80) \) - 0 cookies and 80 muffins
These combinations will allow them to meet their goal of raising at least $300 while selling no more than 80 items.
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