Let's first address how to solve each part of the question in detail.
### (c) Finding the Maximum Revenue
To identify how to find the maximum revenue, you need to understand the behavior of the revenue function [tex]\( R(x) \)[/tex]. Given that [tex]\( R(x) = -x^2 + 6x \)[/tex], you can see that it is a quadratic function opening downwards (because the coefficient of [tex]\( x^2 \)[/tex] is negative). The maximum value of a downward-opening parabola occurs at its vertex.
The necessary step to find this maximum is by determining the vertex of the parabola.
#### How can the maximum revenue be found?
- Option A: Find the maximum [tex]\( y \)[/tex]-coordinate of a point where [tex]\( R(x) = C(x) \)[/tex].
This option is incorrect because it relates to the point where revenue equals cost, not where revenue is maximized.
- Option B: Find the [tex]\( y \)[/tex]-intercept of [tex]\( R(x) \)[/tex].
The [tex]\( y \)[/tex]-intercept is simply where [tex]\( x = 0 \)[/tex], which does not give the maximum value for the quadratic function.
- Option C: Find the [tex]\( y \)[/tex]-coordinate of the vertex of [tex]\( R(x) \)[/tex].
This option is correct because for a quadratic function [tex]\( ax^2 + bx + c \)[/tex] the maximum or minimum value (depending on the opening direction) is at the vertex.
- Option D: Find the [tex]\( x \)[/tex]-coordinate of the vertex of [tex]\( R(x) \)[/tex].
This option is partly correct as it finds where the maximum occurs, but does not give the actual maximum revenue, which is the [tex]\( y \)[/tex]-coordinate of the vertex.
Thus, the correct choice is C. The maximum revenue is the [tex]\( y \)[/tex]-coordinate of the vertex of [tex]\( R(x) \)[/tex], calculated as follows:
[tex]\[ x = -\frac{b}{2a} \][/tex]
where [tex]\( a = -1 \)[/tex] and [tex]\( b = 6 \)[/tex]:
[tex]\[ x = -\frac{6}{2 \times -1} = 3 \][/tex]
Substitute [tex]\( x = 3 \)[/tex] back into [tex]\( R(x) \)[/tex] to find the maximum revenue:
[tex]\[ R(3) = -3^2 + 6 \cdot 3 = -9 + 18 = 9 \text{ (in thousands of dollars)} \][/tex]
[tex]\[ \text{So, the maximum revenue is } 9 \times 1000 = 9000 \text{ dollars.} \][/tex]
### (d) Finding the Maximum Profit
To find the maximum profit [tex]\( P(x) \)[/tex], you first need to define the profit function [tex]\( P(x) \)[/tex] in terms of [tex]\( R(x) \)[/tex] and [tex]\( C(x) \)[/tex]:
[tex]\[ P(x) = R(x) - C(x) \][/tex]
Given:
[tex]\[ R(x) = -x^2 + 6x \][/tex]
[tex]\[ C(x) = x + 4 \][/tex]
The profit function [tex]\( P(x) \)[/tex] becomes:
[tex]\[ P(x) = (-x^2 + 6x) - (x + 4) \][/tex]
[tex]\[ P(x) = -x^2 + 6x - x - 4 \][/tex]
[tex]\[ P(x) = -x^2 + 5x - 4 \][/tex]
To find the maximum profit, again, find the vertex of the quadratic profit function:
[tex]\[ x = -\frac{b}{2a} = -\frac{5}{2 \times -1} = \frac{5}{2} \][/tex]
Substitute [tex]\( x = \frac{5}{2} \)[/tex] back into [tex]\( P(x) \)[/tex]:
[tex]\[ P\left( \frac{5}{2} \right) = -\left( \frac{5}{2} \right)^2 + 5 \cdot \frac{5}{2} - 4 \][/tex]
[tex]\[ P\left( \frac{5}{2} \right) = -\frac{25}{4} + \frac{25}{2} - 4 \][/tex]
[tex]\[ P\left( \frac{5}{2} \right) = -\frac{25}{4} + \frac{50}{4} - \frac{16}{4} \][/tex]
[tex]\[ P\left( \frac{5}{2} \right) = \frac{50 - 25 - 16}{4} \][/tex]
[tex]\[ P\left( \frac{5}{2} \right) = \frac{9}{4} \approx 2.25 \][/tex]
Therefore, the maximum profit is approximately:
[tex]\[ 2 \times 1000 = 2000 \text{ dollars (in thousands) } \][/tex]
So, the maximum profit is [tex]\( 2 \times 1000 = 2000 \)[/tex] dollars.
