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Linear Function

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We determine the extrema of

$\displaystyle f(x,y,z) = x + 2y - z$

subject to the constraints

$\displaystyle x^2 + y^2 - 8 = 0,\quad x+z-4 = 0\,, $

which describe an ellipse as the intersection of a cylinder and a plane.

The Jacobi matrix of the constraints is

$\displaystyle \begin{pmatrix}2x & 2y & 0 \\ 1 & 0 & 1 \end{pmatrix}\,. $

Its rank is maximal unless $ x=y=0$, a case which can be excluded in view of the constraints. Hence, for any extremum $ (x,y,z)$, there exist Lagrange multipliers $ \lambda_1$, $ \lambda_2$ with

$\displaystyle (1, 2 , -1) = (\lambda_1, \lambda_2) \begin{pmatrix}2x & 2y & 0 \\ 1 & 0 & 1 \end{pmatrix} \,, $

i.e.,
$\displaystyle 1$ $\displaystyle =$ $\displaystyle 2\lambda_1 x + \lambda_2$  
$\displaystyle 2$ $\displaystyle =$ $\displaystyle 2\lambda_1 y$  
$\displaystyle -1$ $\displaystyle =$ $\displaystyle \lambda_2 \,.$  

Substituting $ \lambda_1 =1/y$ and $ \lambda_2=-1$ into the first equation, yields $ x = y$. Then, considering the constraints, lead to the points $ (2,2,2)$ and $ (-2,-2,6)$ as possible extrema. Since $ f$ has a minimum as well as a maximum on the ellipse, a comparison of the values of the target function,

$\displaystyle f(-2,-2,6)=-12<4=f(2,2,2)\,, $

shows that $ f$ attains its minimum at $ (-2,-2,6)$ and its maximum at $ (2,2,2)$.

see also:

  automatisch erstellt am 26.  1. 2017