RELATIVE MAXIMUM AND MINIMUM OF MULTI-VARIABLE FUNCTION

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DEFINITION:

1. A function $f\left(x,y\right)$ has a relative minimum at the point $\left(a,b\right)$ if $f\left(x,y\right)\ge f\left(a,b\right)$ for all points$\left(x,y\right)$ in some region around $\left(a,b\right)$.
   
   
2. A function $f\left(x,y\right)$ has a relative maximum at the point $\left(a,b\right)$ if $f\left(x,y\right)\le f\left(a,b\right)$ for all points$\left(x,y\right)$ in some region around $\left(a,b\right)$.

3. The point $\left(a,b\right)$ is a critical point (or a stationary point) of $f\left(x,y\right)$ if  

   $f_x\left(a,b\right)=0$ and $f_y\left(a,b\right)=0.$

WORKING RULE

Suppose that $\left(a,b\right)$ is a critical point of $f\left(x,y\right)$ and that the second order partial derivatives are continuous in some region that contains $\left(a,b\right)$. Next define,

$$D=D\left(a,b\right)=f_{xx}\left(a,b\right)f_{yy}\left(a,b\right)-{\left[f_{xy}\left(a,b\right)\right]}^2$$

We then have the following classifications of the critical point.

1. If $D>0$ and $f_{xx}\left(a,b\right)>0$ then there is a relative minimum at $\left(a,b\right)$.
2. If $D>0$ and $f_{xx}\left(a,b\right)<0$ then there is a relative maximum at $\left(a,b\right)$.
3. If $D<0$ then the point $\left(a,b\right)$ is a saddle point.
4. If $D=0$ then the point $\left(a,b\right)$ may be a relative minimum, relative maximum or a saddle point. Other techniques would need to be used to classify the critical point.

EXAMPLE:  

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

Solution: we will first need to get all the first and second order derivatives.

$$\begin{array}{c}{f}_x=6xy-6x{f}_y=3x^2+3y^2-6y\\ f_{xx}=6y-6f_{yy}=6y-6f_{xy}=6x\end{array}$$

We’ll first need the critical points. for critical point we know 

fx=0=6xy−6x

fy=0=3x2+3y2−6y
so,

$$\begin{array}{cc}6xy-6x& =0\\ 3x^2+3y^2-6y& =0\end{array}$$

These equations are a little trickier to solve than the first set, but once you see what to do they really aren’t terribly bad.

First, let’s notice that we can factor out a 6$x$ from the first equation to get,

$$6x\left(y-1\right)=0$$

So, we can see that the first equation will be zero if $x=0$ or $y=1$. Be careful to not just cancel the $x$ from both sides. If we had done that we would have missed $x=0$.

To find the critical points we can plug these (individually) into the second equation and solve for the remaining variable.

$x=0$ :$$3y^2-6y=3y\left(y-2\right)=0\Rightarrow y=0,y=2$$$y=1$ :$$3x^2-3=3\left(x^2-1\right)=0\Rightarrow x=-1,x=1$$

So, if $x=0$ we have the following critical points,

$$\left(0,0\right)\left(0,2\right)$$

and if $y=1$ the critical points are,

$$\left(1,1\right)\left(-1,1\right)$$

Now all we need to do is classify the critical points. To do this we’ll need the general formula for $D$.

$$D\left(x,y\right)=\left(6y-6\right)\left(6y-6\right)-{\left(6x\right)}^2={\left(6y-6\right)}^2-36x^2$$$\left(0,0\right)$ :$$D=D\left(0,0\right)=36>0f_{xx}\left(0,0\right)=-6<0$$$\left(0,2\right)$ :$$D=D\left(0,2\right)=36>0f_{xx}\left(0,2\right)=6>0$$$\left(1,1\right)$ :$$D=D\left(1,1\right)=-36<0$$$\left(-1,1\right)$ :$$D=D\left(-1,1\right)=-36<0$$

So, it looks like we have the following classification of each of these critical points.

$$\begin{array}{cccc}& \left(0,0\right)& & :\text{Relative Maximum}\\ & \left(0,2\right)& & :\text{Relative Minimum}\\ & \left(1,1\right)& & :\text{Saddle Point}\\ & \left(-1,1\right)& & :\text{Saddle Point}\end{array}$$

HOMEWORK:

Find the critical points for each of the following functions, and use the second derivative test to find the local extrema:

1. f(x,y)=4x2+9y2+8x−36y+24
   
2. g(x,y)=13x3+y2+2xy−6x−3y+4