Quick easy calculus question

[Roy7]

Hey everyone I've got a quick question about easy calculus that I just dont remember. Any help and explanation would be tremendously apprecaited.

The relationship between price and demand fort tickets to an amusement park is given by the equation p = 290-.08q^2 where p is the price of a ticket in dollars and q is the number of people who would buy tickets at that price. What is the maximum revenue the owner can achieve by adjusting the price of the ticket.

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[DrBowtie]

Take the derivative with respect to q and solve for zero for the critical numbers.
This is where the slope will be zero. Plug them back into the original equation. The answer will be the one that produces the most profit.

EDIT: I thought p was profit and q price. I'll have to rethink.

 

[novawildcat]

Revenue is price times quantity so R=pq=q(290-.9q^2)


Find dR/dq set it equal to zero and find the value for q, that maximizes revenue (make sure that this is indeed a maximum). Plug in q into the equation for price and then you are done.

 

[gdbaby]

I didn't know quick, easy and calculus could be used in the same sentence. Bravo.

 

[DesiMcatAcer]

I think you already have your question answered, but one thing I would like to add is you can take the second derivative to check for mximum(or that it exists or not) or the approach suggested by others above

 

[Roy7]

Thanks a million guys I really appreciate it.

 

[potato51]

Calculus is like 7.5 years back for me, but here's what I remember.

Revenue will be p x q = q (290-0.08q^2) = 290q - 0.08q^3.

Take derivative w/ respect to q and set it equal to zero (to see where the function hits a min or max) --> deriv. = 290 - 0.24q^2 = 0

q^2 = 290/0.24... q has to be the positive sq. root to make sense. Plus your answer for 'q' into the formula for revenue and there you go.