Answer to Price to charge to maximize revenue?

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Looking for an answer to this problem:

I understand total revenue calculations, and have tried to find the calculation needed but don't understand.

Question:

Currently, the demand equation for a bicycle is Q = 1000 – 2P. The current price is $150 per bicycle. Is this the best price to charge in order to maximize revenues?*?

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Your revenue is $$ R=p(1000-2p)$$

In order to maximize it you take derivative and you get $$R'=1000-4p$$

So the best price is $$p=250$$ where the revenue is $$R=250(500)=125000$$

With the current price of $150$ your revenue is $$R=150(700)=105000$$