b) Given the demand equation
[tex]\begin{gathered} p=-0.18x+360 \\ p\rightarrow price,x\rightarrow quantity \\ \end{gathered}[/tex]
The revenue function is given by
[tex]\begin{gathered} R(x)=px \\ \Rightarrow R(x)=-0.18x^2+360x \end{gathered}[/tex]
R(x) corresponds to a parabola on the plane that opens downwards; therefore, it has a maximum.
To calculate the maximum, solve the equation R'(x)=0, as shown below
[tex]\begin{gathered} R^{\prime}(x)=-0.18(2x)+360=-0.36x+360 \\ \Rightarrow-0.36x+360=0 \\ \Rightarrow0.36x=360 \\ \Rightarrow x=1000 \end{gathered}[/tex]
The value of x that maximizes the revenue is x=1000.
As for the maximum value of the revenue, find R(1000),
[tex]R(1000)=-0.18(1000)^2+360(1000)=180000[/tex]
The maximum revenue is $180000
c) Find the corresponding price (p) for x=1000,
[tex]p(1000)=-0.18(1000)+360=180[/tex]
The price that maximizes revenue is $180
