Answer:
Explanation:
Derivatives problem! I love these!
To maximize revenue: Find where R(p) has a maximum
1. Find the derivative
-4p^2 + 2000p - 3000
= (2)(-4p) + (1)2000 - (0)3000
= -8p + 2000
2. Find where the derivative = 0
-8p + 2000 = 0
2000 = 8p
250 = p
a. Price = 250
3. Plug in p to your original equation
-4(250)^2 + 2000(250) - 3000
= 247,000
b. Maximum revenue: $247,000