# Math 22 Lagrange Multipliers

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## Method of Lagrange Multipliers

 If ${\displaystyle f(x,y)}$ has a maximum or minimum subject to the constraint ${\displaystyle g(x,y)=0}$, then it will occur at one of the critical numbers of the function  defined by
${\displaystyle F(x,y,\lambda )=f(x,y)-\lambda g(x,y)}$.

In this section, we need to set up the system of equations:

${\displaystyle F_{x}(x,y,\lambda )=0}$
${\displaystyle F_{y}(x,y,\lambda )=0}$
${\displaystyle F_{\lambda }(x,y,\lambda )=0}$


Example: Set up the Lagrange Multipliers:

1) Maximum: ${\displaystyle f(x,y)=xy}$ and Constraint ${\displaystyle x+y-14=0}$

Solution:
So, ${\displaystyle F(x,y,\lambda )=f(x,y)-\lambda g(x,y)=xy-\lambda (x+y-14)=xy-\lambda x-\lambda y+14\lambda }$
${\displaystyle F_{x}(x,y,\lambda )=y-\lambda }$
${\displaystyle F_{y}(x,y,\lambda )=x-\lambda }$
${\displaystyle F_{\lambda }(x,y,\lambda )=-x-y+14}$

2) Maximum: ${\displaystyle f(x,y)=xy}$ and Constraint ${\displaystyle x+3y-6=0}$

Solution:
So, ${\displaystyle F(x,y,\lambda )=f(x,y)-\lambda g(x,y)=xy-\lambda (x+3y-6)=xy-\lambda x-3\lambda y+6\lambda }$
${\displaystyle F_{x}(x,y,\lambda )=y-\lambda }$
${\displaystyle F_{y}(x,y,\lambda )=x-3\lambda }$
${\displaystyle F_{\lambda }(x,y,\lambda )=-x-3y+6}$