1. ## local max, min and saddle points

ok, hello!

Determine all local minimum, maximum and saddle points for $f(x,y)=x^3-y^3-15xy$.

2. Originally Posted by jacek
ok, hello!

Determine all local minimum, maximum and saddle points for $f(x,y)=x^3-y^3-15xy$.
This is not in the form y=f(x)= function of x.
y is part of the equation itself.

What value is this equal to ?
Is it zero? or some constant ?

3. That's whole content of the task.

4. Sorry,
it's a three dimensional function,
not y=function of x,
I haven't done these before...

5. This is not a question of "find the max and min of y as a function of x", it is "find the max and min of f(x,y)". That is, think of z= f(x,y) and find critical points of z.

The critical points of a function of two variables are where $\nabla f= 0$ which is the same as saying [tex]\frac{\partial f}{\partial x}= 0[tex] and $\frac{\partial f}{\partial y}= 0$.

The "second derivative test" for a function of two variables is more complicated than for a function of one variable. First calculate
$\left|\begin{array}{cc}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2}\end{array}\right|= \left(\frac{\partial^2f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right)- \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$.

If D> 0, then the point is a maximum if $\frac{\partial^2 f}{\partial x^2}$ is positive and a minimum if it is negative. If D< 0, then the point is a saddle point.

6. Ah yes! Thanks HallsofIvy,

$Z=f(x,y)=x^3-y^3-15xy$

A critical point occurs when both partial derivatives are zero.

$\frac{\partial f(x,y)}{\partial x}=3x^2-15y=0\ for\ x^2=5y\ \Rightarrow\ y=\frac{x^2}{5}$

$\frac{\partial f(x,y)}{\partial y}=-3y^2-15x=0\ for\ y^2=-5x\ \Rightarrow\ x=-\frac{y^2}{5}$

Hence, we solve the resulting pair of equations for x and y.

$y=\frac{y^4}{125}\ \Rightarrow\ 125y=y^4\ \Rightarrow\ y\left(y^3-125\right)=0\ \Rightarrow\ y=0\ and\ 5$

$y=0\ \Rightarrow\ x=0$
$y=5\ \Rightarrow\ x=-5$

The critical points are (0,0) and (-5,5).

To determine the nature of the critical points, the second partial derivatives are examined.

Also required is $\frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}$

$\frac{\partial^2 f(x,y)}{\partial x^2}=6x$

$\frac{\partial^2 f(x,y)}{\partial y^2}=-6y$

$\frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}=-15$

$D=\frac{\partial^2 f(x,y)}{\partial x^2}\ \frac{\partial^2 f(x,y)}{\partial y^2}-\left(\frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}\right)^2=-36xy-(-15)^2=-36xy-225$

D>0 and $\ \color{blue}\frac{\partial^2 f(x,y)}{\partial x^2}>0\ \Rightarrow\$ relative minimum point

D>0 and $\ \color{blue}\frac{\partial^2 f(x,y)}{\partial x^2}<0\ \Rightarrow\$ relative maximum point

D<0 $\color{blue}\Rightarrow\$ saddle point

At (0,0) D<0 therefore this is a saddle point.

At (-5,5) D is 900-225 >0 and $\frac{\partial^2 f(x,y)}{\partial x^2} = -30 <0$ so this is a relative maximum.

7. on the very end, why do we use only second derivative of x?

8. The 2nd derivative of y is also utilised,
the results for "D", the "discriminant" are determined by the equation for D,
which is regarded as a theorem.

The equation for D does involve the product of both 2nd derivatives.

The saddle points have a "local maximum for x" and a "local minimum for y" colliding at a point or vice versa, which is a way of putting it.
Hence no maximum for both x and y simultaneously (or maximum for both) to identify the point as either a maximum or minimum.

If D>0, then a critical point will be a maximum or minimum.

The reason is based on the formula for D.

$D=\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}-\left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$

Since the RHS of the above expression is always positive,
then the only way D can be >0 is for both 2nd derivatives to be simultaneously positive or simultaneously negative.
Hence, if D>0, there is no need to double-check both 2nd derivatives.

9. Can anyone check if this is correct?

$f(x,y)=6x^2y-12y^2-x^4+18y$

two points:
x=3 y=3
x=-3 y=3

Discriminant is: $-y-5x^2$
And for both points we get -48, so there's only one saddle point and no local max. or min.

10. Originally Posted by jacek
Can anyone check if this is correct?

$f(x,y)=6x^2y-12y^2-x^4+18y$

two points:
x=3 y=3
x=-3 y=3

Discriminant is: $-y-5x^2$
And for both points we get -48, so there's only one saddle point and no local max. or min.

hi jacek,

there are 3 critical points.

$\frac{\partial}{\partial_x}f(x,y)=12xy-4x^3=0\ \Rightarrow\ 3xy-x^3=0\ \Rightarrow\ x\left(3y-x^2\right)=0$

$x=0,\ y=\frac{x^2}{3}$

$\frac{\partial}{\partial_y}f(x,y)=6x^2-24y+18=0\ \Rightarrow\ x^2+3=4y\ \Rightarrow\ y=\frac{x^2+3}{4}$

This yields the 3 critical points (0,0), (-3,3), (3,3).

$\frac{\partial^2}{\partial_x}f(x,y)=12y-12x^2$

$\frac{\partial^2}{\partial_y}f(x,y)=-24$

$\frac{\partial^2}{\partial_x\partial_y}f(x,y)=12x$

$D=12\left(y-x^2\right)(-24)-(12x)^2$

At $(0,0), D=0,$ cannot be identified as a saddle point using the 2nd derivative test.

At $(-3,3)\ D>0,\ f_{xx}<0$ local maximum

At $(3,3)\ D>0,\ f_{xx}<0$ local maximum

11. So, when it comes to discriminant, I cannot cancel 12's from 12y-12x^2 to make it y-x^2?

12. Originally Posted by jacek
So, when it comes to discriminant, I cannot cancel 12's from 12y-12x^2 to make it y-x^2?
No, you can't.

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