\(f(x,y)=y^4+4y^2-x^2\)
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 orthogonal slopes are all zero and 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 $(x,y)$. 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 recognizing 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:
$f_x(x,y)=-2x$ and $f_y(x,y)=4y^3+8y$
For critical points, let:
$-2x=0\implies x=0$
and $4y^3+8y=0\implies 4y(y^2+2)=0$
or $y=0$
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.
Graph of $f(x,y)=y^4+4y^2-x^2$
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
$f_x=0\implies 2x+3y-3=0$ or $2x+3y=3$ (1)
$f_y=0\implies 3x+8y=0$ (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 $\left(\dfrac{24}{7},-\dfrac{9}{7}\right)$.
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