MHB Analyzing Stationary Points of Multivariable Function

[Pulty]

$$f(x,y) = \frac{1}{2}{x}^{2}{e}^{y}-\frac{1}{3}{x}^{3}-y{e}^{3y}$$

To start off I found the partial derivatives of

$$x: {e}^{y}x - {x}^{2}$$

$$y: \frac{{e}^{y}{x}^{2}}{2}-{e}^{3y}-3{e}^{3y}y$$

Then solved simultaneously for each equation equal to 0.

$$y = \ln(x) = -\frac{1}{6}$$

$$x = {e}^{-\frac{1}{6}}$$

(Is there another stationary point?)

I'm unsure of where to go from here, or if I'm even doing the right thing to begin with.

Thank you

 

[MarkFL]

You are correct in that you first want to identify all critical points, by equating the first partials to zero:

$$f_x(x,y)=xe^y-x^2=x\left(e^y-x\right)=0$$

$$f_y(x,y)=\frac{1}{2}x^2e^y-\left(3ye^{3y}+e^{3y}\right)=\frac{1}{2}e^{3y}\left(x^2e^{-2y}-6y-2\right)=0$$

And solving them simultaneously.

From these we obtain:

$$(x,y)=\left(0,-\frac{1}{3}\right),\,\left(e^{-\frac{1}{6}},-\frac{1}{6}\right)$$

So, you had one point, you just missed when $x=0$ from $f_x=0$. :)

Now, armed with these two points, you want to apply the second partials test for relative extrema. Essentially, you want to compute:

$$D(x,y)=f_{xx}(x,y)f_{yy}(x,y)-\left[f_{xy}(x,y)\right]^2$$

What do you get?

 

[Pulty]

fxx(x,y) = e^y - 2x

fyy(x,y) = -9e^(3y)y - 6e^(3y) + (e^(y)+x^(2))/2

fyx(x,y) = e^(y)x

At point (0,-1/3):

fxx(x,y)fyy(x,y) - fxy(x,y)^2 = -0.79079 (Saddle point)

At point (e^1/6, -1/6)

fxx(x,y)fyy(x,y) - fxy(x,y)^2 = 3.08049 (Maximum point)

How does this look?

Also, I am confused as to how you find the point (0,-1/3)

Thanks

 

[MarkFL]

Computing the second partials, we find:

$$f_{xx}(x,y)=e^y-2x$$

$$f_{yy}(x,y)=\frac{1}{2}x^2e^y-\left(\left(9ye^{3y}+3e^{3y}\right)+3e^{3y}\right)=\frac{1}{2}x^2e^y-3e^{3y}\left(3y+2\right)=\frac{1}{2}e^{3y}\left(x^2e^{-2y}-6(3y+2)\right)$$

$$f_{xy}(x,y)=xe^y$$

And so for the first critical point, we have:

$$f_{xx}\left(0,-\frac{1}{3}\right)=e^{-\frac{1}{3}}$$

$$f_{yy}=-\frac{3}{e}$$

$$f_{xy}\left(0,-\frac{1}{3}\right)=0$$

$$D\left(0,-\frac{1}{3}\right)=e^{-\frac{1}{3}}\left(-\frac{3}{e}\right)-0^2=-3e^{-\frac{4}{3}}<0$$

This point is not an extremum.

And for the second critical point, we have:

$$f_{xx}\left(e^{-\frac{1}{6}},-\frac{1}{6}\right)=-e^{-\frac{1}{6}}$$

$$f_{yy}\left(e^{-\frac{1}{6}},-\frac{1}{6}\right)=\frac{1}{2}e^{-\frac{1}{2}}\left(e^{-\frac{2}{3}}-9\right)$$

$$f_{xy}\left(e^{-\frac{1}{6}},-\frac{1}{6}\right)=e^{-\frac{1}{3}}$$

$$D\left(e^{-\frac{1}{6}},-\frac{1}{6}\right)=\left(-e^{-\frac{1}{6}}\right)\left(\frac{1}{2}e^{-\frac{1}{2}}\left(e^{-\frac{2}{3}}-9\right)\right)-\left(e^{-\frac{1}{3}}\right)^2>0$$

This point is a relative maximum.

So, I agree with your conclusions regarding the critical points. :D

How I found the other critical point was from:

$$f_x(0,y)=0$$

And so we then set:

$$f_y(0,y)=-e^{3y}(3y+1)=0\implies y=-\frac{1}{3}$$

And so we find the point:

$$(x,y)=\left(0,-\frac{1}{3}\right)$$

simultaneously satisfies:

$$f_x(x,y)=0$$

$$f_y(x,y)=0$$

as required.

 

[Pulty]

Thanks Mark!

I have a similar question I will ask in a new thread.