Equality Constraints
KKT Conditions
Find the minimum (over $x$, $y$) of the function $f(x,y)$, subject to $g(x,y)=0$, where
\[\begin{alignat*}{2} & \text{minimize: } && 2 x^2 + 3 y^2 \\ & \text{subject to: } && \begin{aligned}[t] x^2 + y^2 &= 4 \end{aligned} \end{alignat*}\]Step I
Defining variable and functions
$\displaystyle
f = 2 x^{2} + 3 y^{2}
g = x^{2} + y^{2} - 4
$
Step II
Defining lagrangian function.
The lagrangian $L=- \lambda \left(x^{2} + y^{2} - 4\right) + 2 x^{2} + 3 y^{2}$
Step III
Deriving KKT equations
$\displaystyle - 2 \lambda x + 4 x= 0 \- 2 \lambda y + 6 y= 0 \x^{2} + y^{2} - 4= 0 \$
Step IV
Solving KKT Conditions to obtain necessary points
| $x$ | $y$ | $\lambda$ | Obj |
|---|---|---|---|
| $-2$ | $0$ | $2$ | $8$ |
| $2$ | $0$ | $2$ | $8$ |
| $0$ | $-2$ | $3$ | $12$ |
| $0$ | $2$ | $3$ | $12$ |
Step V
Computing Bordered Hessian for each points
$\displaystyle \bar{H} = \left[\begin{matrix}0 & 2 x & 2 y\2 x & - 2 \lambda + 4 & 0\2 y & 0 & - 2 \lambda + 6\end{matrix}\right]$
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.