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Homework Help: Finding Max and Min Application of Derivatives

  1. Nov 30, 2007 #1
    Finding Max and Min "Application of Derivatives"

    Hello There!

    The following question I need help in and it would be appreciated. Thank you very much!
    Manufacturing Tires Your company can manufacture x hundred grade A tires and y hundred grade B tires a day where 0 (greater than or equal to) x (less than or equal to) 4 and

    y = (40-10x)/(5-x)

    Your profit on a grade A tire is twice your profit on a grade B tire. What is the most profitable number of each kind to make?

    Thank for your time!
  2. jcsd
  3. Nov 30, 2007 #2


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    Start by writing an expression for your total profit.

    2 side point.
    1. If you read the guidelines you agreed to when you registered you will know that we require you to do your own work.

    2. We have forums specifically for homework. This thread has been moved there.
  4. Nov 30, 2007 #3
    alright, sry about that, I'm just new here!
  5. Nov 30, 2007 #4


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    At a stationary value, the first derivative is zero. At a max value,y''=-ve and at a min value,y''=+ve
  6. Nov 30, 2007 #5
    i found there to be no critical points if I just took the derivative, but I'm not sure if I was correct.
  7. Nov 30, 2007 #6


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    What did you take the derivitive of? Please show us.
  8. Nov 30, 2007 #7
    i just took the derivative of the equation given, as to finding the max when the derivative is set to zero.
  9. Dec 1, 2007 #8


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    The given expression is NOT the profit. You need to think about the profit. That will be the number of items sold times the profit for each. Can you write an expression for the total profit? How about the profit for product A?
  10. Dec 1, 2007 #9
    Since your profit on an A tire is twice as much as the B tire then P = 2A + B. From what is given above, we know that A is the same thing as 100x. We also know that B is the same thing as 100y.You can go ahead and sub those in to the profit equation from above. From here you will want everything to be one variable. You already have a nice relationship between x and y, so another simple substitution there should suffice. Once you do that, use that equation to look for extrema.
  11. Dec 1, 2007 #10
    wow thanks, i didn't see it from that point
  12. Dec 1, 2007 #11
    So this is how i did it, P = 2(100x) + 100y

    and then i subbed y = (40 - 10x)/(5-x) for y in the profit equation, P = 2(100x) + 100y

    I think im misunderstanding what your trying to tell me

    Then i took the derivative and got an "ugly" number whereas should be a whole number for x
  13. Dec 1, 2007 #12


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    One of the things we are trying to tell you, that you are apparently misunderstanding, is "show us what you did". Okay, you substituted y= (40-10x)/(5-x) into P= 2(100x)+ 100y. What did you get? When you differentiated that what did you get?
  14. Dec 1, 2007 #13
    once i subbed it into the the profit equation i took the derivative and got x to be

    (10 (+ or minus) squareroot 80)/ 2

    The x should be a whole number according to the answer.

    276 grade A and 553 grade B tires
  15. Dec 1, 2007 #14
    anyone got any ideas???
  16. Dec 2, 2007 #15
    sry about that i madea mistake in finding the derivative:

    My answer when i took the derivative is x^3 - 10x^2+25x - 5

    When i tried to get the x's I can't get what is the answer. Did i make a mistake
  17. Dec 2, 2007 #16


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    If you will not answer the questions I asked, I cannot help you.
  18. Dec 2, 2007 #17
    what do u mean, do u want me to show it step by step?
  19. Dec 2, 2007 #18
    nevermind i got the answer! Thanks for all your help!
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