R(c)=5000c-100c^2
R'(c)= 5000-200c
Where is R'(c) > 0 and where is R'(c) < 0
What should you charge per download if you want to maximize your total revenue, and how much total revenue will you generate?
+270 I think you may be in my class.. I have this exact same problem on my homework 
Haven't worked it out yet, though.
Yeah I think so too. You should totally help me if you figure it out first LOL
+23252 Since the formula for R'(c) = 5000 - 200c:
To find where R'(x) > 0: 5000 - 200c > 0 ---> -200c > -5000 ---> c < 25
To find where R'(c) < 0: 5000 - 200c < 0 ---> -200c < -5000 ---> c > 25
The maximum occurs where R'(c) = 0: 5000 - 200c = 0 ---> -200c = -5000 ---> c = 25
When c = 25, the revenue will be R(25) = 5000(25) - 100(25)2 = 62,500
