1. Penalty function

How can i minimize
$f(x,y) = 1/2 (x+1) + y$
with Penalty function method

with constraints of
$1-x \leq 0$
$2-y \leq 0$

2. First i set $Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2$

than take the derivatives

$\frac{\partial Q}{\partial x_1}=\frac1{2} +2 \mu (1-x_1) + 2 \mu (2-x_1)$

$\frac{\partial Q}{\partial x_2}=1 +2 \mu (1-x_1) + 2 \mu (2-x_1)$

Then?

Originally Posted by trante
How can i minimize
$f(x,y) = \frac1{2} (x_1+1) + x_2$
with Penalty function method

with constraints of
$1-x_1 \leq 0$
$2-x_2 \leq 0$

3. The function will be
as
$

a) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2
$

or
$b) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1)^2 + (2-x_1)^2]
$

or
$c) Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1 + 2-x_1)^2]
$

???

Originally Posted by trante
First i set $Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu (1-x_1)^2 + \mu (2-x_1)^2$

than take the derivatives

$\frac{\partial Q}{\partial x_1}=\frac1{2} +2 \mu (1-x_1) + 2 \mu (2-x_1)$

$\frac{\partial Q}{\partial x_2}=1 +2 \mu (1-x_1) + 2 \mu (2-x_1)$

Then?

4. $f(x,y) = \frac1{2} (x_1+1) + x_2$
with constraints of
$1-x_1 \leq 0$
$2-x_2 \leq 0$

i set $Q(x,\mu)=\frac1{2} ( x_1+1 ) + x_2 + \mu [ (1-x_1)^2 + (2-x_2)^2]$

$\frac{\partial Q}{\partial x_1}=\frac1{2} +2 \mu (1-x_1)=0$

$\frac{\partial Q}{\partial x_2}=1 +2 \mu (2-x_2)=0$

then $\frac1{2(1-x_1)}=\frac1{2-x_2}$
then $2x_2=x_1$

$2 \mu (2-x_2)=4\mu-2\mu x_2=-1$
$x_2=\frac{4\mu+1}{2\mu}$

$x_2=2$
$x_1=1$

Is it the solution?