SOLUTION: suppose that the manufacturer of a dvd player has found that, when the unit price is p dollars, the revenue r (in dollars) as a function of the price p is r(p)=-2.5p^2+800p. (a)

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Question 1115177: suppose that the manufacturer of a dvd player has found that, when the unit price is p dollars, the revenue r (in dollars) as a function of the price p is r(p)=-2.5p^2+800p.
(a) for what price will the revenue be maximized?
(b) what is the maximum revenue?

Answer by ikleyn(52787)   (Show Source): You can put this solution on YOUR website!
.
The revenue will be maximized when the quadratic function 


r(p) = =-2.5p^2 + 800p 


will get the maximum.


It will happen when  p =  =  = 160 dollars.


It is optimal price for dvd player.


And the maximum revenue then will be  then  -2.5*160^2 + 800*160 = 64000 dollars.


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On finding the maximum/minimum of a quadratic function see the lessons in this site
    - HOW TO complete the square to find the minimum/maximum of a quadratic function
    - Briefly on finding the minimum/maximum of a quadratic function
    - HOW TO complete the square to find the vertex of a parabola
    - Briefly on finding the vertex of a parabola

Also,  you have this free of charge online textbook in ALGEBRA-I in this site
    - ALGEBRA-I - YOUR ONLINE TEXTBOOK.

The referred lessons are the part of this textbook under the topic "Finding minimum/maximum of quadratic functions".


Save the link to this online textbook together with its description

Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson

to your archive and use it when it is needed.



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