The question

so $z = xye^{-(x^2 + y^2)}$

and I need to find the stationary points of the function.

I understand that I need to work out the partial derivative with respect to x, and then equate this term to zero, and then to find which values of x and y this is valid for (i.e. finding the x and y point at which the gradient is zero)

This process is then repeated but with finding the partial derivative with respect to y, equating this to zero and finding which values of x and y this is valid for.

So I calculated both of these partial derivatives and got the correct terms, but I don't understand how the points at which the gradient are zero are found from these partial derivative equations.

The two equations I am left with are:

$$ 0 = (1-2x^2)ye^{-(x^2 + y^2)} $$


$$ 0 = (1-2y^2)xe^{-(x^2 + y^2)} $$

Starting with the first, the e term is brought to the other side (multiplying by zero) so I am left with:

$$ 0 = (1-2x^2)y $$

and repeating this process for the partial derivative with respect to y term:

$$ 0 = (1-2y^2)x $$

From these two equations, how are these results found? :

Stationary points

  • 3
    $\begingroup$ If $ab=0$ then either $a=0$ or $b=0$ (or both). $\endgroup$ – copper.hat Dec 8 '16 at 22:15
  • $\begingroup$ I am sorry that some people have chosen to mark this down. I have marked it up to compensate :) You've clearly worked at this and should be commended for doing so! As @copper says, you have equations where the product is zero. So at least one of the terms in each must be zero. Try all possibilities and you'll get the right answer. $\endgroup$ – Scott Burns Dec 9 '16 at 0:10

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