# Critical points, extrema, and saddle points of $$$f{\left(x,y \right)} = x^{4} - 4 x y + y^{4} + 1$$$

Related calculator: Lagrange Multipliers Calculator

### Your Input

**Find and classify the critical points of $$$f{\left(x,y \right)} = x^{4} - 4 x y + y^{4} + 1$$$.**

### Solution

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

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

$$$\frac{\partial}{\partial y} \left(x^{4} - 4 x y + y^{4} + 1\right) = - 4 x + 4 y^{3}$$$ (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 $$$\begin{cases} 4 x^{3} - 4 y = 0 \\ - 4 x + 4 y^{3} = 0 \end{cases}$$$.

The system has the following real solutions: $$$\left(x, y\right) = \left(-1, -1\right)$$$, $$$\left(x, y\right) = \left(0, 0\right)$$$, $$$\left(x, y\right) = \left(1, 1\right)$$$.

Now, let's try to classify them.

Find all the second-order partial derivatives:

$$$\frac{\partial^{2}}{\partial x^{2}} \left(x^{4} - 4 x y + y^{4} + 1\right) = 12 x^{2}$$$ (for steps, see partial derivative calculator).

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

$$$\frac{\partial^{2}}{\partial y^{2}} \left(x^{4} - 4 x y + y^{4} + 1\right) = 12 y^{2}$$$ (for steps, see partial derivative calculator).

Define the expression $$$D = \frac{\partial ^{2}f}{\partial x^{2}} \frac{\partial ^{2}f}{\partial y^{2}} - \left(\frac{\partial ^{2}f}{\partial y\partial x}\right)^{2} = 144 x^{2} y^{2} - 16.$$$

Since $$$D{\left(-1,-1 \right)} = 128$$$ is greater than $$$0$$$ and $$$\frac{\partial^{2}}{\partial x^{2}} \left(x^{4} - 4 x y + y^{4} + 1\right)|_{\left(\left(x, y\right) = \left(-1, -1\right)\right)} = 12$$$ is greater than $$$0$$$, it can be stated that $$$\left(-1, -1\right)$$$ is a relative minimum.

Since $$$D{\left(0,0 \right)} = -16$$$ is less than $$$0$$$, it can be stated that $$$\left(0, 0\right)$$$ is a saddle point.

Since $$$D{\left(1,1 \right)} = 128$$$ is greater than $$$0$$$ and $$$\frac{\partial^{2}}{\partial x^{2}} \left(x^{4} - 4 x y + y^{4} + 1\right)|_{\left(\left(x, y\right) = \left(1, 1\right)\right)} = 12$$$ is greater than $$$0$$$, it can be stated that $$$\left(1, 1\right)$$$ is a relative minimum.

### Answer

**Relative Maxima**

**No relative maxima.**

**Relative Minima**

**$$$\left(x, y\right) = \left(-1, -1\right)$$$A, $$$f{\left(-1,-1 \right)} = -1$$$A**

**$$$\left(x, y\right) = \left(1, 1\right)$$$A, $$$f{\left(1,1 \right)} = -1$$$A**

**Saddle Points**

**$$$\left(x, y\right) = \left(0, 0\right)$$$A, $$$f{\left(0,0 \right)} = 1$$$A**