Answer:
The price that will maximize the revenue = $10
The maximum revenue = $13,500
Step-by-step explanation:
Given that ⇒ R = -15 p² + 300 p + 12,000
At the maximum revenue
[tex]\frac{dR}{dP} = -30p+300=0[/tex]
Solve for p
∴ -30p +300 = 0
-30 p = -300
p = -300/-30 = 10
So, R = -15 * 10² + 300 * 10 + 12,000 = $13,500
So, The price that will maximize the revenue = $10
and the maximum revenue = $13,500
