# 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*}

### 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

$$x$$ $$y$$ $$\lambda$$ 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

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

### 6. Step VI¶

Determinant of the bordered hessian will provide maxima and minima.

$$x$$ $$y$$ $$\lambda$$ 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.