Question

I keep trying and trying but I cant figure out this problem, please help if you can.

Answer 1

b) Given the demand equation

`\begin{gathered} p=-0.18x+360 \\ p\rightarrow price,x\rightarrow quantity \\ \end{gathered}`

The revenue function is given by

`\begin{gathered} R(x)=px \\ \Rightarrow R(x)=-0.18x^2+360x \end{gathered}`

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

`\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}`

The value of x that maximizes the revenue is x=1000.

As for the maximum value of the revenue, find R(1000),

`R(1000)=-0.18(1000)^2+360(1000)=180000`

The maximum revenue is $180000

c) Find the corresponding price (p) for x=1000,

`p(1000)=-0.18(1000)+360=180`

The price that maximizes revenue is $180