Mathematik-Online lexicon: Linear Function

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	Mathematik-Online lexicon:
	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

, a case which can be excluded in view of the constraints. Hence, for any extremum

, 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

. Then, considering the constraints, lead to the points

and

as possible extrema. Since

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

attains its minimum at

and its maximum at

automatisch erstellt am 26. 1. 2017