# Critical Points, Extrema, and Saddle Points Calculator

The calculator will try to find the critical (stationary) points, the relative (local) maxima and minima, as well as the saddle points of the multivariable function, with steps shown.

Related calculator: Lagrange Multipliers Calculator

## Your Input

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

## Solution

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:

$$$\frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 \left(y - 1\right)$$$ (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.

## Answer

**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**