1. ## Stationary Points

Happy new year guys! Don't ask me why I'm doing maths on New Years Eve

Question:
Find the stationary points and examine the following function for relative extrema and saddle points:
$\displaystyle f(x,y) = 4x^2 e^y - 2x^4 - e^{4y}$

$\displaystyle f_x = 8xe^y - 8x^3 = 0$ which implies x=0 or $\displaystyle x=e^{\frac{y}{2}}$
$\displaystyle f_y = 4x^2e^y - 4e^{4y} = 0$

substituting x=0 gives you end up getting $\displaystyle e^{4y} = 0$ which is not defined. substituting $\displaystyle x=e^{\frac{y}{2}}$ gives you $\displaystyle 4e^{2y}(1-e^{2y}) = 0$ which implies y=0.

So (1,0) is a stationary point - I am able to find the nature of this point, so don't worry about helping me with that.

From the $\displaystyle f_y = 0$ equation, you get $\displaystyle 4e^y (x^2 - e^{3y} ) = 0$, so $\displaystyle x = e^{\frac{3y}{2}}$, substituting into the $\displaystyle f_x$ equation, you get $\displaystyle 8e^{\frac{5y}{2}} - 8e^{\frac{15}{2}}=0$
$\displaystyle y= \frac{2}{5} = 3$

So $\displaystyle ( e^{\frac{9}{2}}, 3)$ is another stationary point. Again I'm able to find the nature of this point.

These seem right because I've substituted them into the $\displaystyle f_x$ and $\displaystyle f_y$ equations and they work.

My concern is there must be more, because looking at the graph, there seem to be at least 2 maximum points and a minimum. Have I gone wrong somewhere or missed something? Thanks for taking the time to look at this.

2. Square roots can be negative!

From the first partial derivative, x=0 or ħe^{y/2}. From the second partial derivative, x=ħe^{3y/2}.

(And don't ask what I'm doing for New Year's Eve!)

3. Originally Posted by Opalg
Square roots can be negative!

From the first partial derivative, x=0 or ħe^{y/2}. From the second partial derivative, x=ħe^{3y/2}.

(And don't ask what I'm doing for New Year's Eve!)
Thank you!!