ok, hello!
Determine all local minimum, maximum and saddle points for $\displaystyle f(x,y)=x^3-y^3-15xy$.
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 $\displaystyle \nabla f= 0$ which is the same as saying [tex]\frac{\partial f}{\partial x}= 0[tex] and $\displaystyle \frac{\partial f}{\partial y}= 0$.
The "second derivative test" for a function of two variables is more complicated than for a function of one variable. First calculate
$\displaystyle \left|\begin{array}{cc}\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x\partial y} \\ \frac{\partial^2 f}{\partial y\partial x} & \frac{\partial^2 f}{\partial y^2}\end{array}\right|= \left(\frac{\partial^2f}{\partial x^2}\right)\left(\frac{\partial^2 f}{\partial y^2}\right)- \left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$.
If D> 0, then the point is a maximum if $\displaystyle \frac{\partial^2 f}{\partial x^2}$ is positive and a minimum if it is negative. If D< 0, then the point is a saddle point.
Ah yes! Thanks HallsofIvy,
$\displaystyle Z=f(x,y)=x^3-y^3-15xy$
A critical point occurs when both partial derivatives are zero.
$\displaystyle \frac{\partial f(x,y)}{\partial x}=3x^2-15y=0\ for\ x^2=5y\ \Rightarrow\ y=\frac{x^2}{5}$
$\displaystyle \frac{\partial f(x,y)}{\partial y}=-3y^2-15x=0\ for\ y^2=-5x\ \Rightarrow\ x=-\frac{y^2}{5}$
Hence, we solve the resulting pair of equations for x and y.
$\displaystyle y=\frac{y^4}{125}\ \Rightarrow\ 125y=y^4\ \Rightarrow\ y\left(y^3-125\right)=0\ \Rightarrow\ y=0\ and\ 5$
$\displaystyle y=0\ \Rightarrow\ x=0$
$\displaystyle y=5\ \Rightarrow\ x=-5$
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 $\displaystyle \frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}$
$\displaystyle \frac{\partial^2 f(x,y)}{\partial x^2}=6x$
$\displaystyle \frac{\partial^2 f(x,y)}{\partial y^2}=-6y$
$\displaystyle \frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}=-15$
$\displaystyle D=\frac{\partial^2 f(x,y)}{\partial x^2}\ \frac{\partial^2 f(x,y)}{\partial y^2}-\left(\frac{\partial}{\partial y}\ \frac{\partial f(x,y)}{\partial x}\right)^2=-36xy-(-15)^2=-36xy-225$
D>0 and $\displaystyle \ \color{blue}\frac{\partial^2 f(x,y)}{\partial x^2}>0\ \Rightarrow\ $ relative minimum point
D>0 and $\displaystyle \ \color{blue}\frac{\partial^2 f(x,y)}{\partial x^2}<0\ \Rightarrow\ $ relative maximum point
D<0 $\displaystyle \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 $\displaystyle \frac{\partial^2 f(x,y)}{\partial x^2} = -30 <0$ so this is a relative maximum.
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.
D<0 for a saddle point.
If D>0, then a critical point will be a maximum or minimum.
The reason is based on the formula for D.
$\displaystyle D=\frac{\partial^2 f}{\partial x^2}\frac{\partial^2 f}{\partial y^2}-\left(\frac{\partial^2 f}{\partial x\partial y}\right)^2$
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.
Can anyone check if this is correct?
$\displaystyle f(x,y)=6x^2y-12y^2-x^4+18y$
two points:
x=3 y=3
x=-3 y=3
Discriminant is: $\displaystyle -y-5x^2$
And for both points we get -48, so there's only one saddle point and no local max. or min.
Please check it!
hi jacek,
there are 3 critical points.
$\displaystyle \frac{\partial}{\partial_x}f(x,y)=12xy-4x^3=0\ \Rightarrow\ 3xy-x^3=0\ \Rightarrow\ x\left(3y-x^2\right)=0$
$\displaystyle x=0,\ y=\frac{x^2}{3}$
$\displaystyle \frac{\partial}{\partial_y}f(x,y)=6x^2-24y+18=0\ \Rightarrow\ x^2+3=4y\ \Rightarrow\ y=\frac{x^2+3}{4}$
This yields the 3 critical points (0,0), (-3,3), (3,3).
$\displaystyle \frac{\partial^2}{\partial_x}f(x,y)=12y-12x^2$
$\displaystyle \frac{\partial^2}{\partial_y}f(x,y)=-24$
$\displaystyle \frac{\partial^2}{\partial_x\partial_y}f(x,y)=12x$
$\displaystyle D=12\left(y-x^2\right)(-24)-(12x)^2$
At $\displaystyle (0,0), D=0,$ cannot be identified as a saddle point using the 2nd derivative test.
At $\displaystyle (-3,3)\ D>0,\ f_{xx}<0$ local maximum
At $\displaystyle (3,3)\ D>0,\ f_{xx}<0 $ local maximum