Question

find all points where the function has any relative extrema or saddle points and identify the type of relative extremum.
f(x,y)=x² + 2y² - xy²
a. relative minimum at (0,0) and saddle points at (2,2) and (2, - 2)
b. saddle point at (0,0) and relative minima at (2,2) and (2, - 2)
c. relative minimum at (0,0) and relative maxima at (2,2) and (2, - 2)
d. relative maximum at (0,0) and saddle points at (2,2) and (2, - 2)

Explanation:

Step1: Find first - order partial derivatives

The function is \(f(x,y)=x^{2}+2y^{2}-xy^{2}\).
The first - order partial derivatives:
\(f_{x}=\frac{\partial f}{\partial x}=2x - y^{2}\)
\(f_{y}=\frac{\partial f}{\partial y}=4y-2xy\)

Step2: Find critical points

Set \(f_{x}=0\) and \(f_{y}=0\).
From \(f_{y}=4y - 2xy=2y(2 - x)=0\), we have two cases:
Case 1: \(y = 0\). Substitute \(y = 0\) into \(f_{x}\): \(2x-y^{2}=2x=0\), so \(x = 0\). The critical point is \((0,0)\).
Case 2: \(x = 2\). Substitute \(x = 2\) into \(f_{x}\): \(2x-y^{2}=4 - y^{2}=0\), then \(y=\pm2\). The critical points are \((2,2)\) and \((2, - 2)\).

Step3: Find second - order partial derivatives

\(f_{xx}=\frac{\partial^{2}f}{\partial x^{2}}=2\), \(f_{xy}=\frac{\partial^{2}f}{\partial x\partial y}=-2y\), \(f_{yy}=\frac{\partial^{2}f}{\partial y^{2}}=4 - 2x\)
The discriminant \(D=f_{xx}f_{yy}-(f_{xy})^{2}\)
For \((0,0)\):
\(f_{xx}(0,0)=2\), \(f_{xy}(0,0)=0\), \(f_{yy}(0,0)=4\)
\(D=(2)(4)-0^{2}=8>0\) and \(f_{xx}(0,0)=2>0\). So \((0,0)\) is a relative minimum.
For \((2,2)\):
\(f_{xx}(2,2)=2\), \(f_{xy}(2,2)=-4\), \(f_{yy}(2,2)=4 - 4 = 0\)
\(D=(2)(0)-(-4)^{2}=-16<0\). So \((2,2)\) is a saddle point.
For \((2,-2)\):
\(f_{xx}(2,-2)=2\), \(f_{xy}(2,-2)=4\), \(f_{yy}(2,-2)=4 - 4 = 0\)
\(D=(2)(0)-(4)^{2}=-16<0\). So \((2,-2)\) is a saddle point.

Answer:

A. Relative minimum at \((0,0)\) and saddle points at \((2,2)\) and \((2, - 2)\)