Solution :
Let the revenue be = R
Therefore, R = price x quantity
R = (30 - 2x) ( 5000 + 500x)
= [tex]150000 + 5000x - 1000x^2[/tex]
For the maximum revenue,
[tex]$\frac{dR}{dx} = 0$[/tex]
[tex]$-2000x+5000=0$[/tex]
[tex]$x=2.5$[/tex]
[tex]$\frac{d^2R}{dx^2}=-2000<0$[/tex]
At [tex]x=2.5[/tex], the revenue is maximum.
So the price for the maximum company revenue = [tex]$30-2x$[/tex]
[tex]$=30-2(2.5)$[/tex]
= 30 - 5
= 25
Therefore, the price that will maximize the company's revenue is $25.
