I'm doing a problem that asks me to find these things for $f(x,y)=y^3+3x^2y-6x^2-6y^2+2$.
I found $f_x=6xy-12x$ and $f_y=3y^2+3x^2-12y$
And found two of the zeros/points listed in the answer key, $(0,0), (0,4)$.
The book says that points $(1,1), (-1,-1)$ are saddle points but I don't see how they got those x and y values as zeros of the second derivatives. In the case of x=1 and y=1, $f_x=6(1)(1)-12(1)=6-12=-6.$
Can someone please explain this?
Finding the Max (downhill in every direction)/Min (uphill in every direction)/Saddle points (down hill along one axis, uphill along the other) for $f(x,y)=y^3+3x^2y-6x^2-6y^2+2$.
Equation to test the points you find to see if they are one of the types you are looking for:
$D$(hessian matrix)$=f_{xx}*f_{yy}-(f_{xy})^2$
-> If D > 0 then the point plugged into its equation is an extrema point. If D < 0 then the point is a saddle point. In the case that D is zero, the test is inconclusive.
-> Need $f_x$ an $f_y$ as well to find the critical points. -> Will use $f+{xx}$ to determine if an extrema point is a maximum point or a minimum point.
Needed partial derivatives:
$f_x=6xy-12x$, $f_y=3y^2+3x^2-12y$
$f_{xx}=6y-12$, $f_{yy}=6y-12$
$f_{xy}=6x$
$D=(6y-12)*(6y-12)-(6x)^2$
Final equation:
$D=36y^2-144y+144$
Finding the critical points (those which are candidates for being a max, min, or saddle point):
->Set $f_x=f_y=0$
$6xy-12x=0$ -----and------ $3y^2+3x^2-12y=0$
-> There is an x or y in every term of both equations. -> Thus setting both x and y to zero sets both equations to zero.
-> First critical point: (0,0)
Now,
-> Use left hand side to find a equation for x.
$6xy-12x=0$ -> $6xy=12x$ -> $\frac{12x}{x}=6y$ -> $12=6y$
-> Thus: y=2. -> Plug y value into right hand side equation.
$3(2^2)+3x^2-12(2)=0$
$12+3x^2-24=0$
$3x^2=12$
-> Thus x = -2 || 2
-> This yields the following critical points: (-2, 2) and (2, 2).
Finally,
-> Notice that setting x to zero, with any y value makes the left hand equation equal to zero. If x equals zero on the right hand side:
$3y^2+3(0^2)-12y$
$(3y^2-12y)=0$
$3(y^2-4y)=0$ -> $y^2-4y=0$
-> Thus y= 4, yielding the following critical point: (0, 4).
-> Therefore the critical points are: (0, 0), (0, 4), (2, 2), (-2, 2).
Testing the points using the formula for D:
$D(0, 0)=36(0)^2-144(0)+144=144$
-> D is positive so this is an extrema point. Plug the point into $f_{xx}$.
$f_{xx}(0,0)=6(0)-12=-12$
-> $f_{xx}$ is negative, the slope in all directions from this point goes downwards, so this is a local max.
$D(0, 4)=36(4)^2-144(4)+144=144$
-> D is positive so this is an extrema point. Plug the point into $f_{xx}$.
$f_{xx}(0,0)=6(4)-12=18$
-> $f_{xx}$ is positive, the slope in all directions from this point goes upwards, so this is a local min.
$D(2, 2)=36(2)^2-144(2)+144=-288$
-> D is negative meaning this is a saddle point.
$D(-2, 2)=36(2)^2-144(2)+144=-288$
-> D is negative meaning this is a saddle point as well.