Dated: 30-10-2024
Examples
Example
\[f(x) = \sqrt{x^2 + y^2}\]
\[f_x(x,y) = \frac{x}{\sqrt{x^2 + y^2}}\]
\[f_y(x,y) = \frac{y}{\sqrt{x^2 + y^2}}\]
The partial derivatives1 exists at all points except for origin.
\[f_x = 0 \text{ only if } x = 0\]
\[f_y = 0 \text{ only if } y = 0\]
The only critical point2 is \((0, 0)\) and \(f(0, 0) = 0\)
Since, \(f(x, y) \ge 0\) for all \((x, y)\), \(f(0, 0) = 0\) is the absolute minimum3 value of \(f\).
Second Partial Derivative1 Test
\[D = f_{xx}(x_0, y_0) \cdot f_{yy}(x_0, y_0) - f^2_{xy}(x_0, y_0)\]
- If \(D > 0\) and \(f_{xx}(x_0, y_0) > 0\) then \(f(x, y)\) has
relative minimum3 at \((x_0, y_0)\) - If \(D > 0\) and \(f_{xx}(x_0, y_0) < 0\) then \(f(x, y)\) has
relative maximum3 at \((x_0, y_0)\) - If \(D < 0\) then \(f(x, y)\) has a
saddle point3 at \((x_0, y_0)\) - If \(D = 0\) then no conclusion can be drawn.
Example
\[f(x, y) = 2x^2 - 4x + xy^2 - 1\]
\[f_x(x, y) = 4x - 4 + y^2 \text{ and } f_{xx}(x, y) = 4\]
\[f_y(x, y) = 2xy \text{ and } f_{yy}(x, y) = 2x\]
\[f_{xy}(x, y) = f_{yx}(x, y) = 2y\]
Putting the partial derivatives1 equal to \(0\), we have
\[4x - 4 + y^2 = 0\]
\[2xy = 0\]
Try solving for variables and we will get following critical points2
\[(1, 0)\]
\[(0, 2)\]
\[(0 ,-2)\]
Checking the nature of the points
\[f_{xx}(1, 0) = 4\]
\[f_{yy}(1, 0) = 2\]
\[f_{xy}(1, 0) = 0\]
\[D = f_{xx}(1, 0) \cdot f_{yy}(1, 0) - f_{xy}^2(1, 0)\]
\[D = 4 \cdot 2 - 0^2 = 8 > 0\]
Hence, the function4 \(f(x, y)\) has relative minimum3 at \((1, 0)\)
\[f_{xx}(0, -2) = 4\]
\[f_{yy}(0, -2) = 0\]
\[f_{xy}(0, -2) = -4\]
\[D = f_{xx}(0, -2) \cdot f_{yy}(0, -2) - f_{xy}^2(0, -2)\]
\[D = 4 \cdot 0 - (-4)^2 = -16 < 0\]
Hence, the function4 \(f(x, y)\) has saddle point3 at \((0, -2)\)
\[f_{xx}(0, 2) = 4\]
\[f_{yy}(0, 2) = 0\]
\[f_{xy}(0, 2) = 4\]
\[D = f_{xx}(0, 2) \cdot f_{yy}(0, 2) - f_{xy}^2(0, 2)\]
\[D = 4 \cdot 0 - 4^2 = -16 < 0\]
Hence, the function4 \(f(x, y)\) has saddle point3 at \((0, 2)\)
References
Read more about notations and symbols.