Equality Constraints

KKT Conditions¶

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

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 \(L=- \lambda \left(x^{2} + y^{2} - 4\right) + 2 x^{2} + 3 y^{2}\)

3. Step III¶

Deriving KKT equations

\(\displaystyle - 2 \lambda x + 4 x= 0 \\- 2 \lambda y + 6 y= 0 \\x^{2} + y^{2} - 4= 0 \\\)

4. Step IV¶

Solving KKT Conditions to obtain necessary points

5. Step V¶

Computing Bordered Hessian for each points

\(\displaystyle \bar{H} = \left[\begin{matrix}0 & 2 x & 2 y\\2 x & - 2 \lambda + 4 & 0\\2 y & 0 & - 2 \lambda + 6\end{matrix}\right]\)

6. Step VI¶

Determinant of the bordered hessian will provide maxima and minima.

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