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Math Help - Extreme Value Theorem

  1. #1
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    Extreme Value Theorem

    Let S be the set of all points inside of the triangle whose vertices are

    (-2,0), (0,2), (2,0).

    Find the minimum and maximum values of the function

    f(x,y) = x^2 + 3y^2 - 4xy + x - y +1

    The equations for the sides of this triangle are:

    y=0 ,       -2 \leq x \leq 2
    y = 2-x ,      0 \leq x \leq 2
    y = 2+x , -2 \leq x \leq 0

    From what I comprehended in class and in my notes, I substitute y into the original function and then take the derivative of the result. I then set the derivative equal to 0 to find the critical points?

    From there I am not too sure what to do...I assume I do it for all three equations and then plug it back into the original function to find the min/max?
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  2. #2
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    Quote Originally Posted by toop View Post
    Let S be the set of all points inside of the triangle whose vertices are

    (-2,0), (0,2), (2,0).

    Find the minimum and maximum values of the function

    f(x,y) = x^2 + 3y^2 - 4xy + x - y +1

    The equations for the sides of this triangle are:

    y=0 , -2 \leq x \leq 2
    y = 2-x , 0 \leq x \leq 2
    y = 2+x , -2 \leq x \leq 0

    From what I comprehended in class and in my notes, I substitute y into the original function and then take the derivative of the result. I then set the derivative equal to 0 to find the critical points?

    From there I am not too sure what to do...I assume I do it for all three equations and then plug it back into the original function to find the min/max?
    this is a quite long exercise! first solve f_x=f_y=0 and choose those solutions that are inside your region. next consider the 3 sides of the triangle and find the values of f in critical points and

    end points of each side. finally points with largest or smallest value of f give you the maximum and minimum of f. to give you a better idea, i'll show you how to do it for one of the sides:

    for example on AB, where A(0,2), \ B(2,0). the equation of the segment AB is: y=2-x, \ 0 \leq x \leq 2. therefore on AB we have:

    f=x^2+3(2-x)^2-4x(2-x)+x - (2-x)+1=8x^2-18x+11, \ 0 \leq x \leq 2. now f'=16x-18=0 gives us: x=\frac{9}{8}, and thus \boxed{f=\frac{7}{8}}. you also need to check the end points, i.e. x = 0 and

    x = 2. at x = 0, we have \boxed{f=11} and at x = 2 we have \boxed{f=7}. now do the same for other two sides.
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  3. #3
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    I appreciate the help!

    When solving for f_x = f_y = 0 I came up with:

    f_x = 2x-4y+1
    f_y = 6y-4x-1

    From there, I am not sure how to set it to 0?

    I tried x = (4y-1) / 2 but not sure where to go from there.


    Also, I understood how you came up with f = 7/8 , f = 11, f = 7 .... do I then just plug it into the original function?

    so: f(9/8,7/8), f(0,11) and so on?
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  4. #4
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    Quote Originally Posted by toop View Post
    I appreciate the help!

    When solving for f_x = f_y = 0 I came up with:

    f_x = 2x-4y+1
    f_y = 6y-4x-1

    From there, I am not sure how to set it to 0?

    I tried x = (4y-1) / 2 but not sure where to go from there.


    Also, I understood how you came up with f = 7/8 , f = 11, f = 7 .... do I then just plug it into the original function?

    so: f(9/8,7/8), f(0,11) and so on?
    no! for x=\frac{9}{8} we get y=\frac{7}{8} and f(\frac{9}{8}, \frac{7}{8})=\frac{7}{8}. similarly for x = 0 and x = 2 we have y = 2 and y = 0 respectively and: f(0,2)=11, \ f(2,0)=7. in order to find the value of f you don't need to

    find the value of y because we already have f as a function of x only. so you only need x. that's why i didn't mention y in my previous post.
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  5. #5
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    After all is said and done...can you verify my answers:



    Min = -5

    Woops, I meant MAX = 11
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  6. #6
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    Quote Originally Posted by toop View Post
    After all is said and done...can you verify my answers:

    Max = 7

    Min = -5
    that's not correct! according to my calculations, which are almost always correct! , the maximum of f is 11 and the minimum is \frac{3}{4}.
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  7. #7
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    Ah, stupid little math errors I made.

    I now have the same answers as you of min=.75 and max=11.

    Thank you very much!
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