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Finding max and min

  1. Dec 8, 2013 #1
    here is the problem plus the solution. I understand all the steps leading to the green box, but within the green box, first of all i don't understand why they put the line in terms of x: (y-6,y) instead of putting it in terms of y: (x,6-x), or does it not matter? and everything else in the box i don't understand why they're doing what they're doing. can someone please explain?


  2. jcsd
  3. Dec 8, 2013 #2


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    Hi iScience! :smile:
    it doesn't matter! :biggrin:

    you can use any parameter (for example, you could use x-y)
    they're expressing f(x,y) in terms of a single parameter (along the hypotenuse), and then finding the max/mins by differentiating that and putting it equal to 0

    (note that the derivative will equal zero at the same point on the hypotenuse, no matter what parameter you differentiate wrt :wink:)

    f = -2(6y2 - y3)

    so df/dy = -2(12y - 3y2), which is zero if y = 0 or 4
  4. Dec 8, 2013 #3
    In single-variable calculus, you learned that the global max/min of a differentiable function on a closed interval had to occur at the endpoints of the interval or at critical points of the interior of the interval. You had to do calculus to identify the critical points, the end points are given to you, and then the problem was just a matter of checking all of the points of interest.

    In the multivariable version, you learn that the global max/min of a differentiable function on a closed set have to occur on the boundary of the set or at critical points of the interior of the set. Considering that the endpoints of an interval constitute the boundary of that interval in the single-variable situation, this is really nothing new conceptually.

    Finding the critical points of the interior involves doing similar calculus as before, however dealing with the boundary in the multivariable case is not as easy. In your problem, there are obvious "pieces" of the boundary, the three sides of the triangle, and so it makes sense to work with each piece individually. On two of the sides, the function is constant. So identifying the max/min of the function on those pieces is easy.

    Identifying the max/min on the third side turns into something the looks like an optimization problem from a first semester of calculus. If you were to go back and look through the section in your first calc class that dealt with optimization, the one that had silly problems about farmers building pens for their cows, you find that most of those problems look a lot like "optimize f(x,y) subject to E(x,y)" where E(x,y) is an equation, sometimes called the constraint equation. The constraint equation was used to reduce a problem in two variables to a problem in one variable. That is exactly what is going on the green box. In this case, the constraint equation is just the equation of the line.
  5. Dec 8, 2013 #4


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    On the line L x+y=6, you can choose either x or y as independent variable. It is y in the green box, so x=6-y. There is a mistake in the writing: f(x,y) is f(6-y, y) on the line L. Otherwise it is correct.

  6. Dec 8, 2013 #5
    Okay i think i got it now. Thanks alot guys!
  7. Dec 8, 2013 #6
    umm, one more thing, if i'm plugging in (6-y,y) into f, i don't get what the solutions got..




    this is a fourth order polynomial.. how did he end up with y^3 as the highest order??..
    Last edited: Dec 8, 2013
  8. Dec 8, 2013 #7
    so the solution is plugging x=(6-y) into an equation of f=-2xy2 but i don't know where this is coming from.

    given in the beginning of the problem is f(x,y) function and of the second term, if we take the derivative of that WRT x then we get "-2xy2" but what about the other terms?..
  9. Dec 9, 2013 #8


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    hi iScience! :smile:
    d'oh! :rolleyes:

    = y2(6-y)(-2) ! :wink:
    f(x,y) = xy2(4 - x - y) everywhere

    on the hypotenuse, that's xy2(4 - 6) :wink:
    not following you :confused:
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