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    henrypoo's Avatar
    henrypoo Posts: 37, Reputation: 0
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    #1

    Mar 11, 2011, 09:34 AM
    find the maximum profit
    May I know how to find the maximum profit and the number of units where the maximum profits occurs. P=-2x^2+72x-198. Can any experts show me how to do it?
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Mar 11, 2011, 09:41 AM

    To find the maximum of a function you:

    1. Find the derivative of the function.
    2. Set the derivative = 0, and find the corresponding values for x.
    3. Evaluate the 2nd derivative of te hfunction at the value of x that you found in step 2. If the 2nd derivative is a negative number, then x corresponds to a local maximum. If the second derivatie is a positive number, then you have a local minimum.
    4. If the domain of the function is limited then you should also check the end points. For your problem you would check the value for p at x = 0 and x = infinity to see if those points are greater than the maximum(s) you found in step 3.

    Here's an example: suppose you have P = -3x^2 + 12x -9. Then:
    1. The derivative is p'(x) = -6x +12.
    2. Solve for p'(x) = 0: -6x +12 = 0, so x = 2,
    3. Evaluate the 2nd derivative at x = 2: p''(x) = -6, which is negative, so x = 2 corresponds to a maximum. p(2) = -3(2)^2 +12(2) -9 = 3.
    4. Check the end points: p(0) = -9, and p(infinity) = - infinity. So p(2) = 3 is indeed a maximum.

    Hope this helps.
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    Unknown008 Posts: 8,076, Reputation: 723
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    #3

    Mar 11, 2011, 09:50 AM

    Just in case you're unfamiliar with those, there is also the completing the square method and one more intuitive method, by factoring.


    This is the latter method:
    If you're plotting the graph of P, you will get a curve, with a maximum.

    You can factor P = -2x^2+72x-198 = -2(x^2 + 36x + 99) = -2(x + 33)(x + 3)

    The curve cuts the x-axis at x = -3 and x = -33, you can get the x coordinate of the curve by finding the midpoint of the two points you just got, then substitute that value in the original equation to get the maximum value of P. From your sketch, you can also see that this point is a maximum.

    I hope it helps! :)
    henrypoo's Avatar
    henrypoo Posts: 37, Reputation: 0
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    #4

    Mar 12, 2011, 02:01 AM
    Here's an example: suppose you have P = -3x^2 + 12x -9. Then:
    1. The derivative is p'(x) = -6x +12.


    how do you get -6x and 12.
    ebaines's Avatar
    ebaines Posts: 12,131, Reputation: 1307
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    #5

    Mar 12, 2011, 06:52 AM
    Quote Originally Posted by henrypoo View Post
    how do you get -6x and 12.
    Not knowing what level of mathematics you are comfortable with, I proposed using a calculus technique. If you're not familiar with derivatives then another way to do this problem is to simply graph the equation P=-2x^2+72x-198 and look for the maximum by eye. Once you have a reasonably good idea of approximately where the max is, try a few values for x around your guess and see if you can zero in on the exact answer.

    I'll give you a hint: the max occurs somewhere between x = 15 and x = 20, and its an integer.
    henrypoo's Avatar
    henrypoo Posts: 37, Reputation: 0
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    #6

    Mar 12, 2011, 10:02 PM
    Is there any website thst can teach me how to draw the graph?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #7

    Mar 13, 2011, 02:49 AM
    Quadratic graphs (those having 2 as the highest power of x) have the general shape of a big 'U'.

    The basic one is



    When you add 1 to it, it moves 1 unit up:



    Similarly, removing 1 moves it down.



    Now, if you add 1 to x, then put the square sign, you move it to the left, or negative direction.



    Similarly, removing 1,



    Remember:
    (x+a)^2 moves the whole graph a units to the left.
    (x-a)^2 moves the whole graph a units to the right.

    Mixing a bit of both now,

    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #8

    Mar 13, 2011, 02:49 AM

    ~ Continued since no more than 6 images can be put in 1 post ~

    And, lastly, putting a number in front of the bracket, or in front of x only changes the shape of the 'U'.



    Putting a negative number overturns the whole graph.



    Note that the equation can also be expressed as:

    And you can see that the graph crosses the x-axis at x = 1 and -3.


    Now, come to your problem, you have

    Can you get an idea of the shape of the graph?
    henrypoo's Avatar
    henrypoo Posts: 37, Reputation: 0
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    #9

    Mar 13, 2011, 04:34 AM
    Sorry I thought is -2(x+33)(x+3)
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #10

    Mar 13, 2011, 04:38 AM

    Yes, I spotted that mistake earlier, and it's supposed really to be -2(x-33)(x-3) in my first post in this thread. Sorry for the confusion about this part.

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