Critical Points, Extrema, and Saddle Points Calculator

Find and classify the critical points of $$$f{\left(x,y \right)} = 2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2$$$.

The first step is to find all the first-order partial derivatives:

$$$\frac{\partial}{\partial x} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 x \left(y - 1\right)$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial}{\partial y} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 2 x^{2} + 3 y^{2} - 4 y$$$ (for steps, see partial derivative calculator).

Next, solve the system $$$\begin{cases} \frac{\partial f}{\partial x} = 0 \\ \frac{\partial f}{\partial y} = 0 \end{cases}$$$, or $$$\begin{cases} 4 x \left(y - 1\right) = 0 \\ 2 x^{2} + 3 y^{2} - 4 y = 0 \end{cases}$$$.

The system has the following real solutions: $$$\left(x, y\right) = \left(0, 0\right)$$$, $$$\left(x, y\right) = \left(0, \frac{4}{3}\right)$$$, $$$\left(x, y\right) = \left(- \frac{\sqrt{2}}{2}, 1\right)$$$, $$$\left(x, y\right) = \left(\frac{\sqrt{2}}{2}, 1\right)$$$.

Now, let's try to classify them.

Find all the second-order partial derivatives:

$$$\frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 y - 4$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial^{2}}{\partial y\partial x} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 x$$$ (for steps, see partial derivative calculator).

$$$\frac{\partial^{2}}{\partial y^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 6 y - 4$$$ (for steps, see partial derivative calculator).

Define the expression $$$D = \frac{\partial ^{2}f}{\partial x^{2}} \frac{\partial ^{2}f}{\partial y^{2}} - \left(\frac{\partial ^{2}f}{\partial y\partial x}\right)^{2} = - 16 x^{2} + 24 y^{2} - 40 y + 16.$$$

Since $$$D{\left(0,0 \right)} = 16$$$ is greater than $$$0$$$ and $$$\frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)|_{\left(\left(x, y\right) = \left(0, 0\right)\right)} = -4$$$ is less than $$$0$$$, it can be stated that $$$\left(0, 0\right)$$$ is a relative maximum.

Since $$$D{\left(0,\frac{4}{3} \right)} = \frac{16}{3}$$$ is greater than $$$0$$$ and $$$\frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)|_{\left(\left(x, y\right) = \left(0, \frac{4}{3}\right)\right)} = \frac{4}{3}$$$ is greater than $$$0$$$, it can be stated that $$$\left(0, \frac{4}{3}\right)$$$ is a relative minimum.

Since $$$D{\left(- \frac{\sqrt{2}}{2},1 \right)} = -8$$$ is less than $$$0$$$, it can be stated that $$$\left(- \frac{\sqrt{2}}{2}, 1\right)$$$ is a saddle point.

Since $$$D{\left(\frac{\sqrt{2}}{2},1 \right)} = -8$$$ is less than $$$0$$$, it can be stated that $$$\left(\frac{\sqrt{2}}{2}, 1\right)$$$ is a saddle point.

Relative Maxima

$$$\left(x, y\right) = \left(0, 0\right)$$$A, $$$f{\left(0,0 \right)} = 2$$$A

Relative Minima

$$$\left(x, y\right) = \left(0, \frac{4}{3}\right)\approx \left(0, 1.333333333333333\right)$$$A, $$$f{\left(0,\frac{4}{3} \right)} = \frac{22}{27}\approx 0.814814814814815$$$A

Saddle Points

$$$\left(x, y\right) = \left(- \frac{\sqrt{2}}{2}, 1\right)\approx \left(-0.707106781186548, 1\right)$$$A, $$$f{\left(- \frac{\sqrt{2}}{2},1 \right)} = 1$$$A

$$$\left(x, y\right) = \left(\frac{\sqrt{2}}{2}, 1\right)\approx \left(0.707106781186548, 1\right)$$$A, $$$f{\left(\frac{\sqrt{2}}{2},1 \right)} = 1$$$A