Certainly! Let's go through the problem step by step to find the revenue function [tex]\( R(x) \)[/tex] and the marginal revenue function [tex]\( R'(x) \)[/tex] for the given demand function.
### Step 1: Define the price function
The price function given is:
[tex]\[ p = 900 - 0.01 x \ln(x) \][/tex]
where [tex]\( x \)[/tex] is the number of yachts and [tex]\( p \)[/tex] is the price per yacht in hundreds of dollars.
### Step 2: Determine the revenue function
The revenue function [tex]\( R(x) \)[/tex] is determined by multiplying the price function [tex]\( p \)[/tex] by the quantity [tex]\( x \)[/tex]. Thus,
[tex]\[ R(x) = p \cdot x \][/tex]
Substitute the price function [tex]\( p \)[/tex] into this equation:
[tex]\[ R(x) = (900 - 0.01 x \ln(x)) \cdot x \][/tex]
[tex]\[ R(x) = 900x - 0.01 x^2 \ln(x) \][/tex]
Therefore, the revenue function is:
[tex]\[ R(x) = 900x - 0.01 x^2 \ln(x) \][/tex]
### Step 3: Find the marginal revenue function
The marginal revenue function [tex]\( R'(x) \)[/tex] is the derivative of the revenue function [tex]\( R(x) \)[/tex] with respect to [tex]\( x \)[/tex].
Given that:
[tex]\[ R(x) = 900x - 0.01 x^2 \ln(x) \][/tex]
We will differentiate [tex]\( R(x) \)[/tex] term-by-term.
First term:
[tex]\[ \text{Derivative of } 900x = 900 \][/tex]
Second term (using the product rule for [tex]\( -0.01x^2 \ln(x) \)[/tex]):
For [tex]\( f(x) = x^2 \)[/tex] and [tex]\( g(x) = \ln(x) \)[/tex]:
[tex]\[
\text{Derivative of } -0.01x^2 \ln(x) = -0.01[(2x \ln(x) + x) = -0.01(2x \ln(x) + x)]
\][/tex]
Therefore,
[tex]\[
R'(x) = 900 - 0.01(2x \ln(x) + x) = 900 - 0.02 x \ln(x) - 0.01 x
\][/tex]
### Conclusion
The revenue function [tex]\( R(x) \)[/tex] and the marginal revenue function [tex]\( R'(x) \)[/tex] for the yacht model are:
[tex]\[ R(x) = 900x - 0.01 x^2 \ln(x) \][/tex]
[tex]\[ R'(x) = 900 - 0.02 x \ln(x) - 0.01 x \][/tex]
