# Finding the local extrema or saddle points of a function

1. Apr 8, 2012

### miglo

1. The problem statement, all variables and given/known data
f(x,y)=5xy-7x^2+3x-6y+2

2. Relevant equations
(f_xx)(f_yy)-(f_xy)^2 the hessian or discriminant of f

3. The attempt at a solution
i arrived at a solution but i dont think its correct, and the answer isnt in the back of the book, so i just wanted to ask if i did this correctly
the first partial derivatives are f_x and f_Y are
f_x=5y-14x+3 and f_y=5x-6
setting f_y=0 i get x=6/5
plugging this value into f_x and solving for y i get 69/25
therefore my critical point is at (6/5,69/25)
the second order partial derivatives are then
f_xx=-14 f_yy=0 and f_xy=5
then using the discriminant of f i get -25 so i get a saddle point
but i graphed the function on wolfram alpha, and i doesnt seem like there is a saddle point on the graph

any help would be greatly appreciated, thanks

2. Apr 8, 2012

### LCKurtz

Are you sure you plotted a region including the purported saddle point?

3. Apr 8, 2012

### Office_Shredder

Staff Emeritus
The problem when you graph it is that the eigenvalues of the Hessian are close to -16 and 1.5. So in the direction where it appears to be a maximum there is a lot of curvature, and in the direction where it appears to be a minimum there is not so much. When you graph it in wolfram alpha there's a clear parabola shape with a ridge at the top and lost due to the scaling is the fact that the ridge does curve up very gently

4. Apr 9, 2012

### miglo

not sure what an eigenvalue is but i guess from both your responses there is a saddle point on f
how do i plot a single region in wolfram alpha? id like to see this saddle point if i can haha

thanks!