Results 1 to 3 of 3

Thread: Maximum and minimum of a function of 2 variables (Lagrange multipliers)

  1. #1
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1

    Maximum and minimum of a function of 2 variables (Lagrange multipliers)

    My question is : Can I solve the problem without passing by Lagrange multipliers? Because I have some difficulties by doing so. If not, then I don't want any tip and I'll work harder.
    I must find the minimum and the maximum of the function $\displaystyle f(x,y)=x^2+y^2+\frac{2xy\sqrt{2}}{3}$ in the ellipse $\displaystyle x^2+2y^2 \leq 1$.
    So it seems a problem that can be solved via Lagrange multipliers, but I didn't find a way to find $\displaystyle \lambda$, $\displaystyle x$ and $\displaystyle y$. (I even had to divide by $\displaystyle 0$ if $\displaystyle \lambda \neq 1$ but then I found another condition ( if $\displaystyle \lambda \neq 0$ then ...) and even with it I couldn't isolate $\displaystyle \lambda$ nor $\displaystyle x$ nor $\displaystyle y$).
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Spec's Avatar
    Joined
    Aug 2007
    Posts
    318
    The requirement that the two gradients are parallel can also be expressed as

    $\displaystyle \left|\begin{array}{cc}f'_x&f'_y\\g'_x&g'_y\end{ar ray}
    \right|=0$

    This will produce exactly the same result as using the Lagrange multipliers method (once you've eliminated the multiplier).

    Both these methods are a bit impractical here though. Since we're dealing with an ellipse, we can express the main function in terms of one variable.

    $\displaystyle f(x,y)=f(\varphi)=\left\{x=\cos\varphi,\ y=\frac{\sin \varphi}{\sqrt{2}} \right\}=\cos^2 \varphi+\frac{\sin^2 \varphi}{2}+\frac{2\cos \varphi\sin \varphi}{3}$

    You will of course also need to find the inner points where $\displaystyle f'_x=f'_y=0$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor arbolis's Avatar
    Joined
    Apr 2008
    From
    Teyateyaneng
    Posts
    1,000
    Awards
    1
    Quote Originally Posted by Spec View Post
    The requirement that the two gradients are parallel can also be expressed as

    $\displaystyle \left|\begin{array}{cc}f'_x&f'_y\\g'_x&g'_y\end{ar ray}
    \right|=0$

    This will produce exactly the same result as using the Lagrange multipliers method (once you've eliminated the multiplier).

    Both these methods are a bit impractical here though. Since we're dealing with an ellipse, we can express the main function in terms of one variable.

    $\displaystyle f(x,y)=f(\varphi)=\left\{x=\cos\varphi,\ y=\frac{\sin \varphi}{\sqrt{2}} \right\}=\cos^2 \varphi+\frac{\sin^2 \varphi}{2}+\frac{2\cos \varphi\sin \varphi}{3}$

    You will of course also need to find the inner points where $\displaystyle f'_x=f'_y=0$
    Thank you VERY much. I'll work on that! (I might ask for further help in a few days)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Maximum and minimum of function
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 9th 2011, 10:54 PM
  2. Replies: 1
    Last Post: Mar 20th 2010, 04:32 AM
  3. Replies: 4
    Last Post: Feb 24th 2010, 02:40 AM
  4. Replies: 0
    Last Post: Feb 23rd 2010, 05:22 PM
  5. Replies: 0
    Last Post: Nov 3rd 2008, 03:49 PM

Search Tags


/mathhelpforum @mathhelpforum