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german
3 years ago
5

Please help I am terrible at algebra

Mathematics
1 answer:
asambeis [7]3 years ago
4 0

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

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