Question

Please help I am terrible at algebra

Answer 1

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

$\frac{dR}{dP} = -30p+300=0$

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