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

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 $f{\left(x,y \right)} = x^{4} - 4 x y + y^{4} + 1$, with steps shown.

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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.

### 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

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