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Non-linear function - Cubic and Quadratic functions

  1. Jul 9, 2012 #1
    1. i am stuck

    2. The profit function is given by P=-2x^2 + 60x - 250, where x units is the quantity and P the total profit. Find the maximum profit and the number of units where maximum profit occurs. Sketch the curve of the profit function.



    3. i cant do anything, my brain just jammed



    ps, sorry, i do not know where to post
     
    Last edited: Jul 9, 2012
  2. jcsd
  3. Jul 9, 2012 #2
    1. i am stuck


    ]2. The profit function is given by P=-2x^2 + 60x - 250, where x units is the quantity and P the total profit. Find the maximum profit and the number of units where maximum profit occurs. Sketch the curve of the profit function.



    3. i started by factorize, but it doesnt work

    PS, i am new, i do not know where to post. sorry
     
    Last edited: Jul 9, 2012
  4. Jul 9, 2012 #3

    Mentallic

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    You found the right place to post :smile:

    Since this is a quadratic function with a negative coefficient of x2 (the -2 is the coefficient) then it'll have a maximum at some x value. The way we can find this x-value and its corresponding P value is by completing the square.
    Do you know how to do that?
     
  5. Jul 9, 2012 #4
    Re: helphelp! i am stuck

    Do you know how to differentiate the function?
     
  6. Jul 9, 2012 #5
    I do not know, i am on private studies, my lecturer just throw a book and assignments to me, and he ask me do it myself, i solve quite a few, left 2 question, thanks to youtube, where else this topic i find it confusing.

    can u help me?
     
  7. Jul 9, 2012 #6
    Re: helphelp! i am stuck

    Then you cannot use differentiation.

    Another method is to try to sketch the graph of P against x. Assume some values for x and find the value of P for each value of x. Then see how the graph goes and try to find the value of x where P attains its maximum value.
     
  8. Jul 9, 2012 #7

    eumyang

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    P = -2x2 + 60x - 250
    It looks like you gave up too soon, my friend. The polynomial is certainly factorable. The first thing one should always try when factoring is to factor out the greatest common factor. What is the greatest common factor in this case?
     
  9. Jul 9, 2012 #8
    Re: helphelp! i am stuck

    is it something like this?:

    -2x^2 + 60x - 250 = P

    -2(x^2 -30x + 125 = P

    this rest i am stuck
     
  10. Jul 9, 2012 #9
    Re: helphelp! i am stuck

    I said ...'Assume some values for x and find the value of P for each value of x'.
     
  11. Jul 9, 2012 #10
    Re: helphelp! i am stuck

    erm....

    p+250=60x-2x^2
    p+250=x(60-2x)
    x must be >1 and <30
    Hence max x is 15.

    p = 15(60-2(15))-250 = 200
     
  12. Jul 9, 2012 #11

    eumyang

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    (missing a parenthesis :wink:)

    Good. Now, what you have inside the parentheses is a trinomial in the form of
    x2 + bx + c

    All you need are two factors of c whose sum is b. For example,
    x2 - 6x - 16 = (x - 8)(x +2)
    The factors of -16 whose sum is -6 are -8 and 2. (-8 x 2 = -16, -8 + 2 = -6)

    In your case,
    P = -2(x2 - 30x + 125)
    you need factors of 125 whose sum is -30. What are the two factors?
     
  13. Jul 9, 2012 #12
    Re: helphelp! i am stuck

    Find the values of the function intersects the x-axis.
    The mid point of intersection is the value of x where P is maximum.
     
  14. Jul 9, 2012 #13
    Re: helphelp! i am stuck

    How did you conclude that x must be >1 and <30?
     
  15. Jul 9, 2012 #14
    -25 and -5 ?
     
  16. Jul 9, 2012 #15
    Thanks! i got it
     
  17. Jul 9, 2012 #16

    Mentallic

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    Umm... Finding the factors does help get you the answer if you know what to do with them when you get there, but completing the square is a direct method that spits out the answer straight away.
     
  18. Jul 9, 2012 #17

    HallsofIvy

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    Re: helphelp! i am stuck

    You are correct that a parabola is symmetric about its vertex. That is, if, for two values of x, the y value is the same number, the vertex is half way between them.

    The standard way to find the maximum or mminimum values for a quadratic (the vertex of the parabola) is to complete the square. Here, we have [itex]y= -2(x^2- 30x+ 125)[/itex].

    A "perfect square" is always of the form [itex](x- a)^2= x^2- 2ax+ a^2[/itex]. Comparing that to our polynomial, we see that we have 2a= -30 so that a= -15 and [itex]a^2= 225[/itex]. [itex]y= -2(x^2- 30x+ 225- 100)= -2(x- 15)^2+ 200= 0[/itex]. When x= 15, x- 15 will be 0 so that y= 200. When x is any number other than 15, x- 15 is non-zero, its square is positive, and we are subtracting it from 200 so y will be less than 200. That is, the maximum value of y is 200 and that happens when x= 15.
     
    Last edited: Jul 9, 2012
  19. Jul 9, 2012 #18

    eumyang

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    Re: helphelp! i am stuck

    The question was reposted here, and reported. OP, please don't repost the same thread in different subforums.
     
  20. Jul 9, 2012 #19

    Ray Vickson

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    The function P(x) = ax^2 + bx + c has (1) a minimum (but no maximum) if a > 0 (its graph looks like that of y = x^2, shifted up or down and left or right); or (2) a maximum (but no minimum) if a < 0 (its graph looks like that of y = -x^2, shifted up or down and left or right).
    [If a = 0 it has neither a maximum nor a minimum if b ≠ 0.]

    So, take the case a < 0, where you want to find the maximum---without using calculus, which you have not taken yet!. What is the largest possible value of P(x) = ax^2 + bx + c? That will be the largest value v for which the equation P(x) = v has a root; if v is too large, it will exceed the maximum of P and the equation will have no root. If v is smaller than the maximum possible value, the equation will have TWO roots. (Just think of plotting the two graphs y = P(x) and y = v, which intersect at two points---the two roots of the equation.)

    How can you identify the maximum? Well, imagine plotting the graph y = P(x) and then imagine moving the line y = v up and up until it just touches the graph of y = P(x) at a single point, say for the value v = v0. If you take v > v0 the two graphs will not intersect at all, so v is larger than the max of P. If v < v0, the two graphs have two intersection points, and so v can be increased a bit and still intersect P; that is, v is less than the maximum of P. Therefore, the maximum of P(x) occurs when the equation P(x) = v has a single root.

    The equation is ax^2 + bx + c = v, or ax^2 + bx + (c-v) = 0. If you solve this using the quadratic formula, you get an expression with a "±" in it, corresponding to the two roots. You have a single root when the expressions have ± 0 in them, and that tells you what must be the relationship between a, b, c and v: you must have b2-4a(c-v)=0. From that you get both the value of v = v0 (the maximum value) and of x (the location of the maximum).

    RGV
     
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