\(f(x,y)=y^4+4y^2-x^2\)

The objective of this question is to find the local minimum and maximum values and the saddle points of the given multi-variable function. For this purpose, a second derivative test is used.

A function of several variables, also known as a real multivariate function, is a function having more than one argument, all of which are real variables. A saddle point is a point on the surface of a function’s graph where the orthogonalÂ slopes are all zero andÂ the function does not have a local extremum.

A point $(x,y)$ on the graph of a function is said to be a local maximum if its $y$ coordinate is greater than all the other $y$ coordinates on the graph at the points near to $(x,y)$.Â More accurately, we can say that $(x,f(x))$ will be a local maximum if $f(x)\geq f(z)$, $x,z\in(a,b)$ and $z\in$ domain of $f$. In a similar way, $(x,y)$ will be a local minimum if $y$ is the smallest coordinate locally, or $(x,f(x))$ will be a local minimum if $f(x)\leq f(z)$, $x,z\in(a,b)$ and $z\in$ domain of $f$.

Local maximum and minimum points on a function graph are quite distinguishable, and thus beneficial in recognizingÂ the shape of the graph.

## Expert Answer

The given function is $f(x,y)=y^4+4y^2-x^2$.

First, find the partial derivatives of the above function as:

$f_x(x,y)=-2x$Â and $f_y(x,y)=4y^3+8y$

For critical points, let:

$-2x=0\implies x=0$

and $4y^3+8y=0\implies 4y(y^2+2)=0$

orÂ $y=0$

Hence, the function has critical points $(x,y)=(0,0)$.

Now for the discriminant $(D)$, we need to find the second order partial partial derivatives as:

$f_{xx}(x,y)=-2$

$f_{yy}(x,y)=12y^2+8$

$f_{xy}(x,y)=0$

And so:

$D=[f_{xx}(x,y)][f_{yy}(x,y)]-[f_{xy}(x,y)]^2$

$D=(-2)(12y^2+8)-(0)^2$

$D=-24y^2-16$

Now at $(0,0)$:

$D=-16$

Therefore, the function has a saddle point at $(0,0)$, and no local maximum or minimum.

## Example

Locate the saddle points, relative minimum or maximum, and the critical points of the function $f$ defined by:

$f(x,y)=x^2+3xy+4y^2-3x$

### Solution

#### Step 1

$f_x=2x+3y-3$

$f_y=3x+8y$

#### Step 2

$f_x=0\implies 2x+3y-3=0$Â orÂ $2x+3y=3$Â Â Â Â Â Â Â Â (1)

$f_y=0\implies 3x+8y=0$Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (2)

Simultaneous solution of (1) and (2) gives us:

$\left(\dfrac{24}{7},-\dfrac{9}{7}\right)$ as a critical point.

#### Step 3

For the discriminant $D$:

$f_{xx}(x,y)=2$

$f_{yy}(x,y)=8$

$f_{xy}(x,y)=3$

$D=[f_{xx}(x,y)][f_{yy}(x,y)]-[f_{xy}(x,y)]^2$

$D=(2)(8)-(3)^2$

$D=7$

Since, $D>0$ and $f_{xx}\left(\dfrac{24}{7},-\dfrac{9}{7}\right)>0$, so by the second derivative test, the function has a local minimum at $\left(\dfrac{24}{7},-\dfrac{9}{7}\right)$.

*Â Images/mathematical drawings are created with GeoGebra.*