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Math Help - i need to find largest revenue

  1. #1
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    i need to find largest revenue

    the demand function given was q= -p(squared) + 33p +9 (18<- p <- 28) copies sold per week where price is p dollars. the question was determine what price the company should charge to obtain largest revenue. they also gave a cost function of c= 9p+100.. they wanted to know largest weekly profit also.

    I tryed to do the rev pq and do derivatives but i kept coming up with squaring a negative..

    Could someone help me out i'd greatly appreciate it!!
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  2. #2
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    Quote Originally Posted by maria View Post
    the demand function given was q= -p(squared) + 33p +9 (18<- p <- 28) copies sold per week where price is p dollars. the question was determine what price the company should charge to obtain largest revenue. they also gave a cost function of c= 9p+100.. they wanted to know largest weekly profit also.

    I tryed to do the rev pq and do derivatives but i kept coming up with squaring a negative..

    Could someone help me out i'd greatly appreciate it!!
    So q is really the quantity sold per week. The product pq will give you the total revenue for price, p. To maximize the revenue, differentiate pq with respect to to q and set it equal to 0.

    You should get in general R'=P+Q\frac{dP}{dQ}. Is this what you were doing?

    Once you solve for that the way to find the weekly profit is pq-c.
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  3. #3
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    more help!

    I am pretty sure I understand, but i could be wrong

    when i did the equation i did q = p(-q(squared) + 33.+9)

    I multiplied p against the demand equation then took the derivative of that and set it equal to zero...

    I ended up getting p(squared) = 33p + 9 but that doesnt seem right?

    any suggestions?
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  4. #4
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    Just in case a picture helps...

    You may as well express R in terms of p then differentiate...

    (I think Jameson meant differentiate with respect to p.)



    Then solve of p when dR/dp = 0 (completing the square) ...

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  5. #5
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    Quote Originally Posted by tom@ballooncalculus View Post

    (I think Jameson meant differentiate with respect to p.)
    Yes you are correct. It seems to me usually the demand or quantity bought is a clear variable while the price is something a company can fully control, so it makes sense to use Q as the variable to differentiate with respect to. Since we were given a way to very nicely get Q for any P we choose, I see why P should be the variable in question.
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  6. #6
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    I really appreciate the help


    The only thing i guess im not getting is

    Is the dr/dp

    what i did was q = -3p(squared) + 66p+9 and set that equal to 0 to solve i think i am missing something in the equation
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  7. #7
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    Quote Originally Posted by maria View Post
    I really appreciate the help


    The only thing i guess im not getting is

    Is the dr/dp

    what i did was q = -3p(squared) + 66p+9 and set that equal to 0 to solve i think i am missing something in the equation
    No that looks correct. You have a quadratic equation in disguise. Think of "p" as "x" and notice you have an p^2 term, a p^1 term and a p^0 term. Use the quadratic formula to solve.

    What you wrote is not Q, it is R', where R is the revenue function. In this problem everything is in terms of price, p, so we use that variable when differentiating. dR/dP is the derivative of the revenue equation with respect to p. You did that already.
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  8. #8
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    THANK YOU!!!

    I understood that completely i really appreciate the help! : )
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  9. #9
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    question again...

    i need some guidance to see if im doing this right

    -p^3 + 33p + 9
    First i did (pq) then i took the derivative to find revenue which i got
    -3p^2 +33p + 9p = 0

    well i keep getting to -p^2 = -11p -3

    is this right? or where do i go from here?

    Also i had to do a cost equation and had to find what -3p^2 + 18p -265 = 0

    I go to p^2= -88.3 + 6p

    Same question here, did i get the right numbers or of i did what do i do to solve?

    thank you : )
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