Equality Constraints

KKT Conditions

Find the minimum (over $x$, $y$) of the function $f(x,y)$, subject to $g(x,y)=0$, where

$$\begin{array}{rlrl}& \text{minimize: }& & 2x^2+3y^2\\ & \text{subject to: }& & \begin{array}{rl}{x}^2+y^2& =4\end{array}\end{array}$$

1. Step I

Defining variable and functions

$\displaystyle f = 2 x^{2} + 3 y^{2}\ g = x^{2} + y^{2} - 4 $

2. Step II

Defining lagrangian function.

The lagrangian $L=-\lambda (x^2+y^2-4)+2x^2+3y^2$

3. Step III

Deriving KKT equations

$$\begin{gathered} {\displaystyle -2\lambda x+4x=0} \\ -2\lambda y+6y=0 \\ {x}^2+y^2-4=0 \end{gathered}$$

4. Step IV

Solving KKT Conditions to obtain necessary points

$x$	$y$	$\lambda$	Obj
$-2$	$0$	$2$	$8$
$2$	$0$	$2$	$8$
$0$	$-2$	$3$	$12$
$0$	$2$	$3$	$12$

5. Step V

Computing Bordered Hessian for each points

${\displaystyle \overline{H}=\left[\begin{array}{ccc}0& 2x& 2y\\ 2x& -2\lambda +4& 0\\ 2y& 0& -2\lambda +6\end{array}\right]}$

6. Step VI

Determinant of the bordered hessian will provide maxima and minima.

$x$	$y$	$\lambda$	Obj	Bordered Hessian
$-2$	$0$	$2$	$8$	$-32$
$2$	$0$	$2$	$8$	$-32$
$0$	$-2$	$3$	$12$	$32$
$0$	$2$	$3$	$12$	$32$

Conclusion: First two points are minima while third and forth points are maxima.

Last modified on: 2023-01-06 22:22:45