a$)$ Take $(x_0,y_0)=(-0\mathrm{.}2,1)$ and find $(x_1,y_1)$ using the following formula:

$(\frac{x_1}{y_1})=(\frac{x_0}{y_0})+$hgradf$(x_0,y_0),$

where $h$ is a small number, say $h=0\mathrm{.}1.$

b$)$ Use the same formula and find $(x_2,y_2).$

If we repeat the procedure, we will find a curve that leads us to the maximum. The graph

is khrown helow

In physics and engineering, we call these curves the gradient flow lines.

c$)$ To find minimum, we should take steps in the opposite direction of gradf. This means that $h$ is a

negative number. The following code in Matlab, gives the minimum point starting at $(1,0\mathrm{.}5).$

f$=$a$($x$,$y$)($x$-$y$.$~$2)**$exp$(-$x$.^2-$y$.$~$2)$;

fx$=$m$($x$,$y$)(1-22+2$y$*2)-\}-2)$;

fy$=$a$($x$,$y$)-2(1+$x$-\backslash$mp@subsup$\{$y$\}\{\}\{\backslash$operatorname$\{$exp$\}(-\backslash$mp@subsup$\{$x$\}\{\}\{-2\}2-\backslash$mp@subsup$\{$y$\}\{\}\{-2\})$;

x$0=1$;y$0=0\mathrm{.}5$; h$=-0\mathrm{.}1$;x$=[$x$0,$y$0]$;

wh$1$le abs $($fx$($x$0,$y$0))+$aba$($fy$($x$0,$y$0))>0\mathrm{.}0001$

x$0=$x$0+$b$*$fx$($x$0,$y$0)$;

y$0=$y$0+$t$**$y$($x$0,$y$0)$;

end

x$=-2$:$0\mathrm{.}1$:$2$;

y$=$x$\text{'}$;

inagesc$($x$,$y$,$f$($x$,$y$),$'alptadata',$0\mathrm{.}5)$

bold on

plot$($X$($:$,1),$X$($:$,2),$'color',$\text{'}$r$\text{'},$'LiseH$1$dth$\text{'},1\mathrm{.}5)$

$[$X $($end$,1),$ X $($end$,2)]$

Run the code and attach the graph to your solution. What is the minimum point of this

procedure?