0

I am trying to find points on $z= (2x^2+y^2)e^(-x^2-y^2)$ where the tangent planes are horizontal. Taking the partial derivatives with respect to x and y and setting both to zero. After simplifying the two equations I get:

$$2=2x^2+3y^2$$ $$3=2x^2+3y^2$$

I know one solution is $(0,0)$, but can anyone help me find the others just by using algebra.

ajotatxe
  • 65,084
Alan
  • 1
  • You should clarify that your simplification is for $x\neq 0$ and $y\neq 0$. Also, the equations should have been $2x^2+y^2=1$ and $2x^2+y^2=2$. Note that by doing your simplification you are losing solutions (where one of $x$ or $y$ is zero and the other is not). You should go through things case by case: $x=0$, then $2x+y^2=2$ etc – smcc Jul 23 '16 at 15:42
  • An equation like $(x/a)^2+(x/b)^2=1$ describes all points on an ellipse. – Maximilian Gerhardt Jul 23 '16 at 15:44

1 Answers1

1

The partial derivatives are zero when

$$2x(2x^2+y^2-2)=0$$ and

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

Go through each case:

You already found $(0,0)$.

Now suppose $x=0$ and $y\neq 0$. Then from the second equation $y^2=1$. This gives two solutions $(0,-1)$ and $(0,1)$.

Next, suppose $x\neq 0$ and $y=0$. Then from the first equation $x^2=1$. This gives two more solutions $(-1,0)$ and $(1,0)$.

Finally, suppose $x\neq 0$ and $y\neq 0$. Then from the two equations:

$$2x^2+y^2=2$$

and

$$2x^2+y^2=1$$

which have no solution (they are contradictory).

smcc
  • 5,694
  • I didn't suppose the conditions. I just saw the two latter equations and got an error. THANK YOU VERY MUCH! – Alan Jul 23 '16 at 15:59