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Maximizing Revenue ahh

  1. Apr 20, 2010 #1
    1. The problem statement, all variables and given/known data

    The Demand Equation for a certian product can be modeled by x+9p=450. Use this to rewrite the revenue equation R=xp as a function of x. Find the number of goods that will maximize profit. Find the maximum profit.

    2. Relevant equations

    x+9p=450
    R=xp

    3. The attempt at a solution
    I have no idea =S. My real problem is i dont know how to write r=xp as a function of x i know it will end up being a quadratic equation i just dont know how to do it. Thanks so much for any help.
     
  2. jcsd
  3. Apr 20, 2010 #2
    Like in my math book examples it takes x=21,000-150p and turns it into R=xp=(21,000-150p)p=-150p^2+21,000p. It says thats expressing the revenue r as a function of p. I dont even understand what they did and thats the first example. I miss 2 days and i get so far behind =(
     
  4. Apr 20, 2010 #3

    Char. Limit

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    Well, first you need to isolate p in your first equation. By this I mean manipulate the equation until you have p, alone, on one side. Then you should have p as a function of x. After this, you can say that p is equal to this function of x, and replace p in your second equation with the function of x.
     
  5. Apr 20, 2010 #4
    so i get p=50-x/9?

    Im sorry but i dont understand your second part.
     
  6. Apr 20, 2010 #5
    oh wait so r=x(50-x/9) right?
     
  7. Apr 20, 2010 #6

    Char. Limit

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    Exactly.

    Now factor all of that out, and you get your quadratic.
     
  8. Apr 21, 2010 #7
    R=50x-x^2/9 is what i got
    well cant i like write it like this R(x)=-1/9x^2+50x

    1/9 is the same as dividing by 9?
     
  9. Apr 21, 2010 #8

    Char. Limit

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    Sounds good.
     
  10. Apr 21, 2010 #9
    alright cool thank you!
     
  11. Apr 21, 2010 #10
    Now to find the number of good that will maximize the profit i take h=-b/2a witch is -50/2(-1/9)= 225.

    I put the 225 in the equation for x like: R(225)=-1/9(225)^2+50(225)= 5626 and thats my maxium profit correct?
     
  12. Apr 21, 2010 #11

    Char. Limit

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    I think there's a typo there, since your math and the correct answer both total to 5625. Maybe you intended to hit that instead?

    If you did, then yes, that is your maximum profit.
     
  13. Apr 21, 2010 #12
    ok haha yeah i did thanks for all the help =)
     
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