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Problem with max and min, i found half solution, need the other half

  1. Dec 24, 2006 #1
    Sorry for bad translation, but data are ok:

    1. The problem statement, all variables and given/known data

    -you have the curve [tex]y = 2x^2 - 4x + 2[/tex]
    -find its intersections with y and x axis, call them A and B.
    -in the parabolic sector limited by these two points, find a point P, in the way that the sum of its cohordinates is minimum and maximum.
    P cohordinates for minimum sum are (3/4 ; 1/8),
    for max sum:(0 ; 2)

    2. Relevant equations

    3. The attempt at a solution

    the general cohordinates of P are P(x ; 2x²-4x+2), so i set the function:
    y=x+2x²-4x+2 and calculate where it has min or maximums, with the condition of [tex]0 \leq x \leq 1[/tex].

    so y'=4x-3: 4x-3 > 0, x > 3/4. now i should put in the disequation the condition [tex]0 \leq x \leq 1[/tex] right? but it goes wrong.

    it gives me a maximum in x=0, ok, a minimum in x=3/4, ok, and another maximum in x=1, not ok.
    for me it's the first attempt to those problems please tell me where i'm wrong.
    (i didn't put cohordinates as absolute values because i saw that in that "x" interval they are ok as i put them).
  2. jcsd
  3. Dec 25, 2006 #2
    can you explain how you got a maximum at 1?
    you get a min point in 3/4 cuz the of the first derivative changes sign there.
    then you check x=0 and x=1 to see if there are maximum points there. f(x) is bigger then f(1) so there is a max at 0 but not at 1.
  4. Dec 25, 2006 #3
    i think i understood, but, what's wrong if i set the sistem of disequations and solve it? please view this picture:i get a max in 1

    http://img98.imageshack.us/img98/743/derivss3.th.png [Broken]
    Last edited by a moderator: May 2, 2017
  5. Dec 25, 2006 #4


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    The intersection with the x axis is, of course, (0, 2) and the intersection with the y axis is where y= 2x2-4x+ 2= 0. That is the same as
    x2- 2x+ 1= (x- 1)2 so, yes, the intersection with the y axis is (1, 0). That means you are looking for the maximum and minimum values of x+ y= x+ 2x2- 4x+ 2= 2x2- 3x+ 2 on the interval [itex]0\le x\le 1[/itex]. The derivative is 4x- 3 and that equals 0 only when x= 3/4.

    Now, the theorem is "Every continuous function has both maximum and minimum values on a closed, bounded interval. If the function is differentiable, those values may occur in the interior of the interval, where the derivative is 0, or at the endpoints of the interval.

    To find the absolute maximum and minimum of this function on the interval [itex]0\le x\le 1[/itex], evaluate the function at x= 0, 3/4, and 1.
  6. Dec 25, 2006 #5
    take a look at this picture:
    http://img143.imageshack.us/img143/3936/graphfq9.th.jpg [Broken]

    it shows the function and the derivative.
    the picture you posted is wrong. nether the function nor the derivetive change direction at 0 or 1.

    when you wanna find min/max points do the following:
    1. find the derivitive of the function
    2. find any points where the derivitive changes sign. these are the local min/max points. check their value.
    3. check the value of the function at the limits of your range

    in your case the derivitive changes sign at only one point. so thats (3/4, 7/8).
    in the limits of your range there is (0,2) and (1, 1).
    clearly the lowest point is (3/4, 7/8) and the highest is (0,2).
    put those values in your original function and you get (0,2) and (3/4, 1/8)
    Last edited by a moderator: May 2, 2017
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