$f (x, y)$ is given by $f (x, y) = (x^2 - 5x\cdot y)\cdot e^y$

These are the question to that function that I'm struggling with:

1. Find the partial derivatives of first and second order of $f(x, y)$.
2. Find the stationary points of $f(x, y)$ and determines for each point on it/they are a local maximum point, the local minimum point or saddle point.
3. Is it possible to say something about the function has maximum and minimum values ​​based on the information you have found?

I've tried over and over and I'm getting real frustrated. It's a bonus problem that I really don't have to do, but I'd like to anyway.

What I got on first problem:

First order: $f'_x(x,y) = (2x-5y)\cdot e^y$ and $f'_y(x,y)= -5x\cdot e^y$.

Correct?

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$(u\times v)' = u' \times v + u \times v'$ so your second one is false. You considered $e^y$ as a constant in both whereas it isn't one in the second. – xavierm02 Oct 7 '12 at 12:48
the usual notation is just $f_x$ and $f_y$. Look up Hessian and second derivative test. – James S. Cook Oct 7 '12 at 12:50
(1) If you can compute the first-order partial derivatives, what prevents you from computing the second? (2) Hint: There's a connection between $(x,y)$ being a stationary point and the values of $F_x(x,y)$ and $F_y(x,y)$. And to decide whether the stationary points are local minima, maxima or saddle points, well, you haven't computed the second order derivates in vain... (3) Hint: In which cases does a function have local minima or maxima but not a global one? And in which cases does it have a global maxima or minima which isn't a local one? – fgp Oct 7 '12 at 12:54
@fgp Thanks for the hints. What prevents me is my intelligence I guess, this was a bonus problem for the smarter guys, I'm not one of them but I'd like to come closer to that level at least. I'll try again with the help I've gotten in here. – Frank Len Oct 7 '12 at 13:56
@FrankLen If you're smart enough to derive once, you're quite probably smart enough to derive twice ;-) Just do what you did to the original function, just start with $F_x$ (and then $F_y$) instead of the original $f$. You're going to get 4 second-order partial derivatives. Btw, these things are easiest to understand if you interpret them geometrically. Make drawings of local minima, maxima and saddle points, and figure out how partial derivatives enter that picture. – fgp Oct 7 '12 at 14:02

(1) It looks like your $f_x$ is correct (noting that one in general do not write $f_x'$ for the derivative. The subscript $x$ shows that you have taken the derivative with respect to $x$) However, your $f_y$ doesn't look quite right.

You have: $$f(x,y) = (x^2 - 5xy)e^y = x^2e^y - 5xye^y$$ So $$f_y = x^2e^y-5xe^y - 5xye^y \quad\text{(product rule).}$$ (2) To find the stationary points you need to solve the system of equations \begin{align} f_x(x,y) = 0 \quad &\text{and}\quad f_y(x,y) = 0.\\ 2x = 5y \quad&\text{and}\quad x^2-5x-5xy = 0\Rightarrow \\ x^2 - 5x - 5x(\frac{2}{5}x) &= 0 \Rightarrow \\ x(x -7) &= 0. \end{align} You can probably solve this...

(3) To classify the stationary points you compute the second order partial derivatives: $$f_{xx}, f_{yy}, f_{xy}, f_{yx}$$ Then you compute the "discriminant": $$D = f_{xx}f_{yy} - f_{xy}f_{yx}.$$ at the stationary points. Then you have \begin{align} D > 0 \text{ and } f_{xx} > 0 &\Rightarrow \text{local minimum} \\ D > 0 \text{ and } f_{xx} < 0 &\Rightarrow \text{local maximum} \\ D < 0 &\Rightarrow \text{saddle point}. \end{align} If $D = 0$, you don't know.

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Thank you, I'll try this. But on (2), I don't quite understand the text after "the system of equations", is it supposed to be pasted into a program? Excel? – Frank Len Oct 7 '12 at 13:43
@FrankLen It's latex code which is how you type in equations here (Well, obviously there must be a different way to type in equation here too, given that you didn't recognize it as latex code, yet your question contains equations...). He forgot the closing $$, which is why the latex code doesn't get rendered. – fgp Oct 7 '12 at 14:05 @FrankLen: Sorry about that. I had indeed forgotten the$$ – Thomas Oct 7 '12 at 14:57
I managed to fy, but what do you mean by this: "f(x,y)=(x^2−5xy)ey=x^2−5xyey"? Wouldn't you have to multiply e^y into x^2 aswell? – Frank Len Oct 7 '12 at 15:20
@FrankLen: I made a mistake. Sorry! I will edit to correct. – Thomas Oct 7 '12 at 18:39