4.8.1 Examples for finding extrema

  1. \begin{eqnarray*}f(x,y)&=&x^2+y^2\\ \rightarrow\,\hat H(x,y)&=&\left(\begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array}\right)\to\mbox{pos.\ definite, i.e.\ minimum}\end{eqnarray*}

  2. \begin{eqnarray*}f(x,y)&=&-x^2-y^2\\ \rightarrow\,\hat H(x,y)&=&\left(\begin{array}{cc} -2 & 0 \\ 0 & -2 \end{array}\right)\to\mbox{neg.\ definite, i.e.\ maximum}\end{eqnarray*}

  3. \begin{eqnarray*}f(x,y)&=&x^2-y^2\\ \rightarrow\,\hat H(x,y)&=&\left(\begin{array}{cc} 2 & 0 \\ 0 & -2 \end{array}\right)\to\mbox{indefinite, i.e.\ no extremum, saddle point}\end{eqnarray*}

  4. semi-cases difficult:

    \[f(x,y)=x^2+y^4\to\vec\nabla f=\vect{2x\\4y^3}\to \hat H(x,y)=\left(\begin{array}{cc} 2 & 0 \\ 0 & 12y^2 \end{array}\right)\]

    \(\hat H\) pos. semi-definite?\(\to\)minimum

    \[f(x,y)=x^2\to\vec\nabla f=\vect{2x\\0}\to \hat H(x,y)=\left(\begin{array}{cc} 2 & 0 \\ 0 & 0 \end{array} \right)\]

    \(\hat H\) pos. semi-definite?\(\to\)mall values \(x=0\) y-arbitrary are minima

    \[f(x,y)=x^2+y^3\to\vec\nabla f=\vect{2x\\3y^2}\to \hat H(x,y)=\left(\begin{array}{cc} 2 & 0 \\ 0 & 6y \end{array} \right)\]

    \(\hat H(0,0)\) pos. semi-definite?\(\to\)but no extremum, complicated saddle point!

  5. \[f(x_1,x_2)=e^{-(x_1^2+x_2^2-4x_1x_2-2x_1)}\quad \vec\nabla f=e^{-(x_1^2+x_2^2-4x_1x_2-2x_1)}\vect{-2x_1+4x_2+2\\-2x_2+4x_1}\]
    \[\vec\nabla f=-2e^{-(x_1^2+x_2^2-4x_1x_2-2x_1)}\vect{2x_2-x_1-1\\2x_1-x_2}\begin{array}{cclcclcc} \rightarrow&2x_2-x_1&=&1\\ \to&2x_1-x_2&=&0\\ &x_2&=&2x_1&\to&4x_1-x_1&=&-1\\ &&&&&x_1&=&-\frac{1}{3}\\ &&&&&x_2&=&-\frac{2}{3}\end{array}\]

    \begin{eqnarray*}\tilde H(x_1,x_2)&=&2e^{-(x_1^2+x_2^2-5x_1x_2-2x_1)}*\\ && \left(\begin{array}{cc}-1+2(2x_2-x_1+1)^2&\quad 2+2(2x_1-x_2)(2x_2-x_1+1)\\2+2(2x_1-x_2)(2x_2-x_1+1)\quad &-1+2(2x_1-x_2)^2\end{array}\right)\end{eqnarray*}

    Ignoring the positive prefactor we find

    \[\tilde H(-\frac{1}{3},-\frac{2}{3})\propto\left(\begin{array}{cc} -1 & 2 \\ 2 & -1 \end{array}\right),\]

    i.e. the Hesse matrix is negative-definite implicating that the extremum at \((-\frac{1}{3},-\frac{2}{3})\) is a maximum.


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© J. Carstensen (Math for MS)