Question
What is the maximum profit Phillip can achieve and the corresponding profit per bottle if the first 9600 bottles fetch $5 profit per bottle, and each additional bottle reduces profit by $0.0002?
Answers
GPT-4o mini
To determine the maximum profit Phillip can achieve, we need to model the profit based on the number of bottles sold.
Let's denote the number of bottles sold by \( x \).
- For the first 9600 bottles, the profit per bottle is $5.
- For each additional bottle beyond 9600, the profit per bottle decreases by $0.0002.
Thus, we need to set up the profit function:
1. For \( x \leq 9600 \):
\[
\text{Profit} = 5x
\]
2. For \( x > 9600 \):
If \( x = 9600 + y \), where \( y \) is the number of extra bottles sold (with \( y \geq 0 \)):
\[
\text{Profit per bottle} = 5 - 0.0002y
\]
The total number of bottles sold is \( x = 9600 + y \), so we can express the profit as:
\[
\text{Profit} = (5 - 0.0002y)(9600 + y)
\]
Expanding this gives:
\[
\text{Profit} = (5 - 0.0002y)(9600) + (5 - 0.0002y)y
\]
\[
\text{Profit} = 48000 - 0.0002 \cdot 9600y + 5y - 0.0002y^2
\]
\[
\text{Profit} = 48000 - 1.92y + 5y - 0.0002y^2
\]
\[
\text{Profit} = 48000 + (5 - 1.92)y - 0.0002y^2
\]
\[
\text{Profit} = 48000 + 3.08y - 0.0002y^2
\]
Now, we need to maximize this quadratic function (which opens downwards due to the negative coefficient of \( y^2 \)).
The maximum value of a quadratic \( ax^2 + bx + c \) occurs at
\[
y = -\frac{b}{2a}
\]
In our case:
- \( a = -0.0002 \)
- \( b = 3.08 \)
Calculating \( y \):
\[
y = -\frac{3.08}{2 \times -0.0002} = \frac{3.08}{0.0004} = 7700
\]
Thus, the maximum number of bottles sold \( x \) when \( y = 7700 \):
\[
x = 9600 + 7700 = 17300
\]
Now, we can calculate the maximum profit:
Substituting \( y = 7700 \) back into the profit function for \( x > 9600 \):
\[
\text{Profit} = 48000 + 3.08 \cdot 7700 - 0.0002 \cdot (7700)^2
\]
Calculating each term:
- \( 3.08 \cdot 7700 = 23616 \)
- \( (7700)^2 = 59290000 \)
- \( 0.0002 \cdot 59290000 = 11858 \)
Putting this together:
\[
\text{Profit} = 48000 + 23616 - 11858
\]
\[
\text{Profit} = 48000 + 11758
\]
\[
\text{Profit} = 59758
\]
Now, we also need to find the profit per bottle when \( x = 17300 \):
The profit per bottle for \( x > 9600 \):
\[
\text{Profit per bottle} = 5 - 0.0002 \cdot 7700 = 5 - 1.54 = 3.46
\]
**Final Results:**
- Maximum Profit: \( \boxed{59758} \)
- Profit per Bottle: \( \boxed{3.46} \)
Let's denote the number of bottles sold by \( x \).
- For the first 9600 bottles, the profit per bottle is $5.
- For each additional bottle beyond 9600, the profit per bottle decreases by $0.0002.
Thus, we need to set up the profit function:
1. For \( x \leq 9600 \):
\[
\text{Profit} = 5x
\]
2. For \( x > 9600 \):
If \( x = 9600 + y \), where \( y \) is the number of extra bottles sold (with \( y \geq 0 \)):
\[
\text{Profit per bottle} = 5 - 0.0002y
\]
The total number of bottles sold is \( x = 9600 + y \), so we can express the profit as:
\[
\text{Profit} = (5 - 0.0002y)(9600 + y)
\]
Expanding this gives:
\[
\text{Profit} = (5 - 0.0002y)(9600) + (5 - 0.0002y)y
\]
\[
\text{Profit} = 48000 - 0.0002 \cdot 9600y + 5y - 0.0002y^2
\]
\[
\text{Profit} = 48000 - 1.92y + 5y - 0.0002y^2
\]
\[
\text{Profit} = 48000 + (5 - 1.92)y - 0.0002y^2
\]
\[
\text{Profit} = 48000 + 3.08y - 0.0002y^2
\]
Now, we need to maximize this quadratic function (which opens downwards due to the negative coefficient of \( y^2 \)).
The maximum value of a quadratic \( ax^2 + bx + c \) occurs at
\[
y = -\frac{b}{2a}
\]
In our case:
- \( a = -0.0002 \)
- \( b = 3.08 \)
Calculating \( y \):
\[
y = -\frac{3.08}{2 \times -0.0002} = \frac{3.08}{0.0004} = 7700
\]
Thus, the maximum number of bottles sold \( x \) when \( y = 7700 \):
\[
x = 9600 + 7700 = 17300
\]
Now, we can calculate the maximum profit:
Substituting \( y = 7700 \) back into the profit function for \( x > 9600 \):
\[
\text{Profit} = 48000 + 3.08 \cdot 7700 - 0.0002 \cdot (7700)^2
\]
Calculating each term:
- \( 3.08 \cdot 7700 = 23616 \)
- \( (7700)^2 = 59290000 \)
- \( 0.0002 \cdot 59290000 = 11858 \)
Putting this together:
\[
\text{Profit} = 48000 + 23616 - 11858
\]
\[
\text{Profit} = 48000 + 11758
\]
\[
\text{Profit} = 59758
\]
Now, we also need to find the profit per bottle when \( x = 17300 \):
The profit per bottle for \( x > 9600 \):
\[
\text{Profit per bottle} = 5 - 0.0002 \cdot 7700 = 5 - 1.54 = 3.46
\]
**Final Results:**
- Maximum Profit: \( \boxed{59758} \)
- Profit per Bottle: \( \boxed{3.46} \)