Results 1 to 7 of 7

Thread: Extreme Value Theorem

  1. #1
    Junior Member
    Joined
    May 2007
    Posts
    36

    Extreme Value Theorem

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

    $\displaystyle (-2,0), (0,2), (2,0).$

    Find the minimum and maximum values of the function

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

    The equations for the sides of this triangle are:

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

    From what I comprehended in class and in my notes, I substitute $\displaystyle y$ 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?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by toop View Post
    Let S be the set of all points inside of the triangle whose vertices are

    $\displaystyle (-2,0), (0,2), (2,0).$

    Find the minimum and maximum values of the function

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

    The equations for the sides of this triangle are:

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

    From what I comprehended in class and in my notes, I substitute $\displaystyle y$ 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 $\displaystyle f_x=f_y=0$ and choose those solutions that are inside your region. next consider the 3 sides of the triangle and find the values of $\displaystyle f$ in critical points and

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

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

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

    x = 2. at x = 0, we have $\displaystyle \boxed{f=11}$ and at x = 2 we have $\displaystyle \boxed{f=7}$. now do the same for other two sides.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    May 2007
    Posts
    36
    I appreciate the help!

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

    $\displaystyle f_x = 2x-4y+1$
    $\displaystyle f_y = 6y-4x-1$

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

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


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

    so: $\displaystyle f(9/8,7/8), f(0,11)$ and so on?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    Quote Originally Posted by toop View Post
    I appreciate the help!

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

    $\displaystyle f_x = 2x-4y+1$
    $\displaystyle f_y = 6y-4x-1$

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

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


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

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

    find the value of y because we already have $\displaystyle f$ as a function of $\displaystyle x$ only. so you only need $\displaystyle x.$ that's why i didn't mention y in my previous post.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2007
    Posts
    36
    After all is said and done...can you verify my answers:



    Min = -5

    Woops, I meant MAX = 11
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    May 2008
    Posts
    2,295
    Thanks
    7
    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 $\displaystyle f$ is $\displaystyle 11$ and the minimum is $\displaystyle \frac{3}{4}.$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    May 2007
    Posts
    36
    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!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Proof: Extreme Value Theorem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Feb 24th 2011, 06:29 PM
  2. Extreme Value Theorem
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: Nov 8th 2010, 05:34 AM
  3. extreme points and choquet theorem
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Jul 16th 2010, 02:45 AM
  4. Extreme Value Theorem
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: Mar 27th 2010, 06:04 PM
  5. Applying the Extreme Value Theorem
    Posted in the Calculus Forum
    Replies: 9
    Last Post: Feb 9th 2008, 03:14 PM

Search Tags


/mathhelpforum @mathhelpforum