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Quadrant by the x and y axes

  1. Jun 23, 2012 #1
    1. The problem statement, all variables and given/known data
    A right triangle is formed in the first quadrant by the x and y axes and a line through the point (1,2). Find the vertices of the triangle such that its area is a minimum.


    2. Relevant equations
    [itex]a=\frac{1}{2}xy[/itex]
    [itex]x^2+y^2=h^2[/itex]
    [itex]y-2=m(x-1)[/itex]
    [itex]p=x+y+h[/itex]
    [itex]p=x+m(x+1)+2+h[/itex]
    [itex]x^2+(mx-m+2)^2=h^2[/itex]

    3. The attempt at a solution
    The last several equations represent my attempt so far. I cannot figure out an equation to differentiate. I also tried trig functions, but that didn't help.
     
  2. jcsd
  3. Jun 24, 2012 #2
    Re: optimization

    Does this actually have a numerical answer? As far as I can see, you have two fixed points, the origin and the point (1,2), which means bringing the other point closer to the origin you decrease the area, and as long as it's not (0,0) it will remain a triangle.
     
  4. Jun 24, 2012 #3
    Re: optimization

    You have to many variables that are not independent. You should be able to focus on one and only one, for example the slope of the the hypotenuse, then write area as a function of the slope and find a's extrema.


     
  5. Jun 24, 2012 #4

    Dick

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    Re: optimization

    Concentrate on the second equation. Find the x and y intercepts in terms of m. Now use the first equation. Differentiate with respect to m to find the extermum.
     
  6. Jun 24, 2012 #5
    Re: optimization

    [itex]x int: (m[x-1]+2, 0)[/itex]
    [itex]y int: (0, 2-m)[/itex]

    how can I get rid of the x in the x intercept

    edit: I can have [itex]x=\frac{2a}{y}[/itex], but doesn't seem to help.

    am going to bed now, so will check in tomorrow, thanks for the assistance.
     
    Last edited: Jun 24, 2012
  7. Jun 24, 2012 #6

    SammyS

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    Re: optimization

    The x-coordinate of x-intercept is equal to the base of the triangle.

    The y-coordinate of y-intercept is equal to the altitude of the triangle.

    How is the area of a triangle related to its base & altitude?
     
  8. Jun 24, 2012 #7

    Dick

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    Re: optimization

    Put y=0 and solve for x to get the x-intercept. The result shouldn't contain an extra 'x'.
     
  9. Jun 24, 2012 #8
    Re: optimization

    Yes, I was really short on sleep when I wrote that. :zzz:

    The equation is [itex]\frac{1}{2}(2-m)(\frac{m-2}{m})=A[/itex]

    I've got it now, thanks.
     
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