Find the stationary points of the following functions, and
identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):
(a) f(x, y) = x2 + y2 - x - y
(b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4
Find the stationary points of the following functions, and
identify the nature of each (ie local maxima, minima, saddle point or point of inflexion):
(a) f(x, y) = x2 + y2 - x - y
(b) g(x, y) = 3x2 - 3xy + 4y2 + 8x - 5y - 4
Hello,
Find $\displaystyle x_0$ and $\displaystyle y_0$ such that $\displaystyle \frac{\partial f}{\partial x}(x_0,y_0)=0$ and $\displaystyle \frac{\partial f}{\partial y}(x_0,y_0)=0$
These are the critical points.
I hope you've done bits of algebra including matrices...
The Hessian matrix is defined at a point as :
$\displaystyle H_f(x_0,y_0)=\begin{pmatrix} \frac{\partial^2 f}{\partial x^2}(x_0,y_0) & \frac{\partial^2 f}{\partial x \partial y}(x_0,y_0) \\ \\ \frac{\partial^2 f}{\partial y \partial x}(x_0,y_0) & \frac{\partial^2 f}{\partial y^2}(x_0,y_0) \end{pmatrix}$
$\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial x}\right) \quad \quad \quad \quad \frac{\partial^2 f}{\partial x \partial y}=\frac{\partial}{\partial x} \left(\frac{\partial f}{\partial y}\right)$
Work with eigenvalues
If the eigenvalues are all positive, then this point $\displaystyle (x_0,y_0)$ is a maximum.
If they are all negative, then this point $\displaystyle (x_0,y_0)$ is a minimum.
If one is positive and one is negative, then this point $\displaystyle (x_0,y_0)$ is a saddle point.
Work with discriminant
If the discriminant is negative, then it's a saddle point.
If the discriminant is positive and $\displaystyle \frac{\partial^2 f}{\partial x^2}(x_0,y_0) >0$, then it's a maximum.
If the discriminant is positive and $\displaystyle \frac{\partial^2 f}{\partial x^2}(x_0,y_0) <0$, then it's a minimum.
If you don't know how to proceed, google for Hessian matrix.
First find where $\displaystyle \frac{\partial f}{\partial x}=0$ and $\displaystyle \frac{\partial f}{\partial y}=0$ [and in part (b), where the partials of g are equal to zero.]
These will give you the critical points. (Be careful on how you pair them together)
Then apply the second partials test to see if its a max, min, or is a saddle point.
$\displaystyle D=f_{xx}(x_0,y_0)f_{yy}(x_0,y_0)-f_{xy}^2(x_0,y_0)$
Recall that if:
$\displaystyle D>0~\text{and}~f_{xx}(x_0,y_0)>0$, then $\displaystyle f$ has a relative minimum at $\displaystyle (x_0,y_0)$
$\displaystyle D>0~\text{and}~f_{xx}(x_0,y_0)<0$, then $\displaystyle f$ has a relative maximum at $\displaystyle (x_0,y_0)$
$\displaystyle D<0$, then $\displaystyle f$ has a saddle point at $\displaystyle (x_0,y_0)$
$\displaystyle D=0$, then no conclusion can be drawn.
You have enough information to help you get started on the problem(s)
--Chris
I'm fairly certain that points of inflection are not considered stationary points. That being said, given f(x, y), you must calculate all points at which df/dx = df/dy = 0, and then you must calculate $\displaystyle f_{xx}, f_{xy}, f_{yx}, f_{yy}$ and evaluate $\displaystyle f_{xx}f_{yy} - f_{xy}f_{yx}$. If that value is positive and both $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ are positive, it means that the point is a minimum. If that value is positive and both $\displaystyle f_{xx}$ and $\displaystyle f_{yy}$ are negative, it means that the point is a maximum. If $\displaystyle f_{xx}f_{yy} - f_{xy}f_{yx}$ is negative, the point is a saddle point. Edited for correctness.
(a): $\displaystyle \frac{df}{dx} = 2x - 1; \frac{df}{dy} = 2y - 1; f_{xx} = 2; f_{yy} = 2; f_{xy} = 0; f_{yx} = 0$
Hence we have x = 0.5; y = 0.5 and thus our only point of consideration is (0.5, 0.5). What type of point is this? Use the formula:
$\displaystyle f_{xx}f_{yy} - f_{xy}f_{yx} = 2 \cdot 2 - 0 \cdot 0 = 4 - 0 = 4$
Since 4 is positive and both 2 and 2 are positive, this point is a local minimum. In this case, it is also an absolute minimum of the function.
See if you can handle (b).
Edited for clarity and correctness.
If you need an example, we need only do one of these, right?
the rules are, if $\displaystyle (a,b)$ is a critical point of a function $\displaystyle f(x,y)$, that is, $\displaystyle f_x(a,b) = f_y(a,b) = 0$ (and of course we assume $\displaystyle f$ is differentiable on some disk containing $\displaystyle (a,b)$ and has continuous partial derivatives on the said disk etc), then we can classify the critical points as follows:
Let $\displaystyle D = D(a,b) = f_{xx}(a,b)f_{yy}(a,b) - [f_{xy}]^2$, then
(a) If $\displaystyle D > 0$ and $\displaystyle f_{xx}>0$, then $\displaystyle f(a,b)$ is a local minimum
(b) If $\displaystyle D > 0$ and $\displaystyle f_{xx}<0$, then $\displaystyle f(a,b)$ is a local maximum
(c) If $\displaystyle D<0$, then $\displaystyle f(a,b)$ is a saddle point.
lets see how this works with your $\displaystyle f(x,y)$
Note that:
$\displaystyle \begin{array}{lcl} f_x = 2x - 1 & & f_y = 2y - 1 \\ & & \\ f_{xx} = 2 & & f_{yy} = 2 \\ & f_{xy} = 0 & \end{array}$
so that $\displaystyle x = y = \frac 12$ are the only values of $\displaystyle x$ and $\displaystyle y$ that make our partials zero. hence, the only critical point is $\displaystyle \bigg( \frac 12 , \frac 12 \bigg)$
Now, $\displaystyle D \bigg( \frac 12 , \frac 12 \bigg) = 4 > 0$ and $\displaystyle f_{xx} \bigg( \frac 12 , \frac 12 \bigg) = 2 > 0$, so that $\displaystyle f \bigg( \frac 12 , \frac 12 \bigg)$ is a local minimum
EDIT: I should have known iceman would pick up the slack
you must be able to at least find the partial derivatives. if not, then us going through these examples will not help you.
please review finding partial derivatives.
to start you off: when you find the derivative of a function with respect to one variable, you treat all other variables as constants. until you can do this, it makes no sense to try these problems. you are essentially trying to run before you even know how to crawl