Question

(2) find the critical points of $f(x,y)=x^{2}-x^{4}+xy + y^{2}$.

Explanation:

Step1: Calculate partial - derivatives

First, find the first - order partial derivatives of \(f(x,y)\).
The partial derivative with respect to \(x\) is:
\[f_x=\frac{\partial f}{\partial x}=2x - 4x^{3}+y\]
The partial derivative with respect to \(y\) is:
\[f_y=\frac{\partial f}{\partial y}=x + 2y\]

Step2: Set partial - derivatives equal to zero

Set \(f_x = 0\) and \(f_y=0\). From \(f_y=x + 2y=0\), we get \(x=-2y\).
Substitute \(x = - 2y\) into \(f_x\):
\[2(-2y)-4(-2y)^{3}+y = 0\]
\[ - 4y+32y^{3}+y = 0\]
\[32y^{3}-3y = 0\]
Factor out \(y\): \(y(32y^{2}-3)=0\)

Step3: Solve for \(y\)

Case 1: If \(y = 0\), then from \(x=-2y\), \(x = 0\).
Case 2: If \(32y^{2}-3=0\), then \(y^{2}=\frac{3}{32}\), so \(y=\pm\frac{\sqrt{3}}{4\sqrt{2}}=\pm\frac{\sqrt{6}}{8}\)
When \(y=\frac{\sqrt{6}}{8}\), \(x=-2y=-\frac{\sqrt{6}}{4}\)
When \(y =-\frac{\sqrt{6}}{8}\), \(x=-2y=\frac{\sqrt{6}}{4}\)

Answer:

The critical points are \((0,0)\), \((-\frac{\sqrt{6}}{4},\frac{\sqrt{6}}{8})\), and \((\frac{\sqrt{6}}{4},-\frac{\sqrt{6}}{8})\)