Question
A reasonable scale for the horizontal axis of the function A(d)=d(250−2d) is %0D%0A, A reasonable scale for the vertical axis of the function A(d)=d(250−2d) is
Answers
GPT-4o mini
To find a reasonable scale for the horizontal and vertical axes for the function \( A(d) = d(250 - 2d) \), we should first analyze the function.
### Step 1: Understanding the Function
The function \( A(d) \) represents a quadratic function. It can be rewritten as follows:
\[
A(d) = 250d - 2d^2
\]
This quadratic opens downwards (since the coefficient of \( d^2 \) is negative), and it will have a maximum value at its vertex.
### Step 2: Finding the Vertex
To find the vertex, we can use the formula for the vertex of a quadratic function \( d = -\frac{b}{2a} \):
Here, \( a = -2 \) and \( b = 250 \):
\[
d = -\frac{250}{2(-2)} = \frac{250}{4} = 62.5
\]
### Step 3: Finding the Maximum Area
To find the maximum value of \( A \):
Plugging \( d = 62.5 \) back into the function:
\[
A(62.5) = 62.5(250 - 2 \times 62.5) = 62.5(250 - 125) = 62.5 \times 125 = 7812.5
\]
### Step 4: Reasonable Axes Scales
- **Horizontal Axis (d-axis) Scale**:
- Since \( d \) represents some parameter (for example, width, length, etc.), and we found that the maximum occurs at \( d = 62.5 \), a reasonable range for \( d \) would be from \( 0 \) to about \( 125 \) (a bit beyond the vertex to see the entire shape).
- **Suggested scale**: 0 to 125 (increments of 25 or 50).
- **Vertical Axis (A(d)-axis) Scale**:
- The maximum area calculated is \( 7812.5 \), so the vertical axis should realistically extend from \( 0 \) to around \( 8000 \).
- **Suggested scale**: 0 to 8000 (increments of 1000 or 2000).
### Conclusion
Based on the analysis, a reasonable scale for the axes can be described as follows:
- **Horizontal Axis (d):** 0 to 125 (increments of 25 or 50)
- **Vertical Axis (A(d)):** 0 to 8000 (increments of 1000 or 2000)
This will give a clear representation of the function and its maximum area.
### Step 1: Understanding the Function
The function \( A(d) \) represents a quadratic function. It can be rewritten as follows:
\[
A(d) = 250d - 2d^2
\]
This quadratic opens downwards (since the coefficient of \( d^2 \) is negative), and it will have a maximum value at its vertex.
### Step 2: Finding the Vertex
To find the vertex, we can use the formula for the vertex of a quadratic function \( d = -\frac{b}{2a} \):
Here, \( a = -2 \) and \( b = 250 \):
\[
d = -\frac{250}{2(-2)} = \frac{250}{4} = 62.5
\]
### Step 3: Finding the Maximum Area
To find the maximum value of \( A \):
Plugging \( d = 62.5 \) back into the function:
\[
A(62.5) = 62.5(250 - 2 \times 62.5) = 62.5(250 - 125) = 62.5 \times 125 = 7812.5
\]
### Step 4: Reasonable Axes Scales
- **Horizontal Axis (d-axis) Scale**:
- Since \( d \) represents some parameter (for example, width, length, etc.), and we found that the maximum occurs at \( d = 62.5 \), a reasonable range for \( d \) would be from \( 0 \) to about \( 125 \) (a bit beyond the vertex to see the entire shape).
- **Suggested scale**: 0 to 125 (increments of 25 or 50).
- **Vertical Axis (A(d)-axis) Scale**:
- The maximum area calculated is \( 7812.5 \), so the vertical axis should realistically extend from \( 0 \) to around \( 8000 \).
- **Suggested scale**: 0 to 8000 (increments of 1000 or 2000).
### Conclusion
Based on the analysis, a reasonable scale for the axes can be described as follows:
- **Horizontal Axis (d):** 0 to 125 (increments of 25 or 50)
- **Vertical Axis (A(d)):** 0 to 8000 (increments of 1000 or 2000)
This will give a clear representation of the function and its maximum area.