Question
The cost, in dollars, to produce x designer dog leashes is C(x)=5x+2, and the revenue function, in dollars, is R(x)=4x^2+101x
Find the profit function.
P(x)=
Find the number of leashes which need to be sold to maximize the profit.
leashes
dogs
dollars
Find the maximum profit.
leashes
dogs
dollars
Find the price to charge per leash to maximize profit.
dogs
leashes
dollars
What would be the best reasons to either pay or not pay that much for a leash?
Find the profit function.
P(x)=
Find the number of leashes which need to be sold to maximize the profit.
leashes
dogs
dollars
Find the maximum profit.
leashes
dogs
dollars
Find the price to charge per leash to maximize profit.
dogs
leashes
dollars
What would be the best reasons to either pay or not pay that much for a leash?
Answers
GPT-4o mini
To solve the problem, we will first determine the profit function \( P(x) \).
1. **Finding the Profit Function \( P(x) \)**:
The profit function is given by the difference between revenue and cost:
\[
P(x) = R(x) - C(x)
\]
Given:
\[
C(x) = 5x + 2
\]
\[
R(x) = 4x^2 + 101x
\]
Thus, we can substitute into the profit function:
\[
P(x) = (4x^2 + 101x) - (5x + 2)
\]
Simplifying this gives:
\[
P(x) = 4x^2 + 101x - 5x - 2
\]
\[
P(x) = 4x^2 + 96x - 2
\]
2. **Finding the Number of Leashes to Maximize Profit**:
To find the maximum profit, we can take the derivative of the profit function and set it to zero:
\[
P'(x) = 8x + 96
\]
Setting the derivative to zero:
\[
8x + 96 = 0
\]
Solving for \( x \):
\[
8x = -96
\]
\[
x = -12
\]
Since the number of leashes produced cannot be negative, we need to check the second derivative to confirm the nature of this critical point:
\[
P''(x) = 8
\]
Because \( P''(x) > 0 \), this indicates that \( P(x) \) has a global minimum at \( x = -12 \). In the context of this problem, the profit maximization on the positive side of the x-axis will occur at the endpoints of our feasible region or through numerical methods (if available). Thus, let's reconsider by finding the vertex of the quadratic - the formula is given by \( x = -\frac{b}{2a} \):
\[
x = -\frac{96}{2 \cdot 4} = -\frac{96}{8} = -12
\]
This implies the calculation of the revenue may need a range check. If revenue increases up into our x range and reflects positivity across our budget based upon point assessment at intervals.
3. **Finding the Maximum Profit**:
We can determine the vertex's x and calculate the maximum profit \( P(x) \) by substituting the found x values in our function, typically we'd calculate around 0 to verify slight endpoints based on cost factors.
4. **Finding the Price to Charge per Leash to Maximize Profit**:
To find the price per leash, we need to find \( \frac{R(x)}{x} \) for the corresponding x values.
5. **Evaluating the Investment in Leash Pricing**:
- Factors to Decide Whether to Pay Up:
- **Quality**: If leashes provide high durability, aesthetic appeal, or have unique features (like an environment-friendly approach), they might justify a higher cost.
- **Brand Reputation**: Established brands may engender consumer trust regarding safety and quality.
- **Market Comparison**: Competitor price points versus function cost, creating a comparative baseline and consumer acceptance threshold.
Conclusively, while your particular leashes might have peaks in maximizing demand at certain costs, evaluating the local market through customer feedback would provide decision support based on affordability to perceived value, making sizing optimizations necessary.
1. **Finding the Profit Function \( P(x) \)**:
The profit function is given by the difference between revenue and cost:
\[
P(x) = R(x) - C(x)
\]
Given:
\[
C(x) = 5x + 2
\]
\[
R(x) = 4x^2 + 101x
\]
Thus, we can substitute into the profit function:
\[
P(x) = (4x^2 + 101x) - (5x + 2)
\]
Simplifying this gives:
\[
P(x) = 4x^2 + 101x - 5x - 2
\]
\[
P(x) = 4x^2 + 96x - 2
\]
2. **Finding the Number of Leashes to Maximize Profit**:
To find the maximum profit, we can take the derivative of the profit function and set it to zero:
\[
P'(x) = 8x + 96
\]
Setting the derivative to zero:
\[
8x + 96 = 0
\]
Solving for \( x \):
\[
8x = -96
\]
\[
x = -12
\]
Since the number of leashes produced cannot be negative, we need to check the second derivative to confirm the nature of this critical point:
\[
P''(x) = 8
\]
Because \( P''(x) > 0 \), this indicates that \( P(x) \) has a global minimum at \( x = -12 \). In the context of this problem, the profit maximization on the positive side of the x-axis will occur at the endpoints of our feasible region or through numerical methods (if available). Thus, let's reconsider by finding the vertex of the quadratic - the formula is given by \( x = -\frac{b}{2a} \):
\[
x = -\frac{96}{2 \cdot 4} = -\frac{96}{8} = -12
\]
This implies the calculation of the revenue may need a range check. If revenue increases up into our x range and reflects positivity across our budget based upon point assessment at intervals.
3. **Finding the Maximum Profit**:
We can determine the vertex's x and calculate the maximum profit \( P(x) \) by substituting the found x values in our function, typically we'd calculate around 0 to verify slight endpoints based on cost factors.
4. **Finding the Price to Charge per Leash to Maximize Profit**:
To find the price per leash, we need to find \( \frac{R(x)}{x} \) for the corresponding x values.
5. **Evaluating the Investment in Leash Pricing**:
- Factors to Decide Whether to Pay Up:
- **Quality**: If leashes provide high durability, aesthetic appeal, or have unique features (like an environment-friendly approach), they might justify a higher cost.
- **Brand Reputation**: Established brands may engender consumer trust regarding safety and quality.
- **Market Comparison**: Competitor price points versus function cost, creating a comparative baseline and consumer acceptance threshold.
Conclusively, while your particular leashes might have peaks in maximizing demand at certain costs, evaluating the local market through customer feedback would provide decision support based on affordability to perceived value, making sizing optimizations necessary.