Examining Critical Points

Question

Investigate the nature of critical points for the given functions:
a) $f(x,y)=x^3-3x^2+y^2$
b) $f(x,y)=x^2+xy+y^2+1/x+1/y$

Solution

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a) $f(x,y)=x^3-3x^2+y^2$

Step-1: Find critical points:

$\begin{align*} \left\ \frac{\partial f}{\partial x} = 3x^2-6x \end{align*}$

$\begin{align*} \left\ \frac{\partial f}{\partial y} = 2y \end{align*}$

$\begin{align*} \left \frac{\partial f}{\partial x} = 0 \ \Rightarrow 3x^2-6x=0 \ \Rightarrow x=2 \end{align*}$

$\begin{align*} \left \frac{\partial f}{\partial y} = 0 \ \Rightarrow 2y=0 \ \Rightarrow y=0 \end{align*}$

Thus, (2, 0) is the critical point for f(x, y).

Step-2: Find Eigenvalues at critical point:

Hessian matrix of f,

$Hf(x,y) = \begin{vmatrix} \frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} \\ \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} \end{vmatrix} = \begin{vmatrix} 6x-6 & 0\\ 0 & 2 \end{vmatrix}$

$Hf(2,0) = \begin{bmatrix} -6 & 0\\ 0 & 2 \end{bmatrix}$

$\mid Hf(2,0) - \lambda I \mid = 0 \Rightarrow \mid Hf(2,0) - \lambda I \mid = 0 \Rightarrow$

$\lambda^2 + 4λ + 12=0 \Rightarrow \lambda = 6, 2$

Step-3: Find nature of the critical point:

Since eigenvalues of Hf(2,0) are positive, (2, 0) is a local minimum.
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