2
$\begingroup$

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?

  • 0
    Thanks again, I guess its just that I find everything related to "e" difficult. Yes, like Fx,Fy and then on the second-order: Fxx,Fyy,Fxy, and Fyx right? My book isn't very good so I'm using examples I can find in here. Btw: Why does it say 1 Answer and then several smaller answers above like yours?2012-10-07

1 Answers 1

2

(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.

  • 0
    @FrankLen: I made a mistake. Sorry! I will edit to correct.2012-10-07