Analyzing Stationary Points of Multivariable Function

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Appreciate your help!In summary, we found the partial derivatives of f(x,y) and solved them simultaneously to find two critical points: (0,-1/3) and (e^-1/6, -1/6). By using the second partials test, we determined that (e^-1/6, -1/6) is a relative maximum while (0,-1/3) is not an extremum. We also found (0,-1/3) by setting f_y(0,y)=0 and solving for y.
  • #1
Pulty
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\(\displaystyle f(x,y) = \frac{1}{2}{x}^{2}{e}^{y}-\frac{1}{3}{x}^{3}-y{e}^{3y}\)

To start off I found the partial derivatives of

\(\displaystyle x: {e}^{y}x - {x}^{2}\)

\(\displaystyle y: \frac{{e}^{y}{x}^{2}}{2}-{e}^{3y}-3{e}^{3y}y\)

Then solved simultaneously for each equation equal to 0.

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

\(\displaystyle x = {e}^{-\frac{1}{6}}\)

(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
 
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  • #2
You are correct in that you first want to identify all critical points, by equating the first partials to zero:

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

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

And solving them simultaneously.

From these we obtain:

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

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:

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

What do you get?
 
  • #3
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
 
  • #4
Computing the second partials, we find:

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

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

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

And so for the first critical point, we have:

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

\(\displaystyle f_{yy}=-\frac{3}{e}\)

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

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

This point is not an extremum.

And for the second critical point, we have:

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

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

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

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

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:

\(\displaystyle f_x(0,y)=0\)

And so we then set:

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

And so we find the point:

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

simultaneously satisfies:

\(\displaystyle f_x(x,y)=0\)

\(\displaystyle f_y(x,y)=0\)

as required.
 
  • #5
Thanks Mark!

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

FAQ: Analyzing Stationary Points of Multivariable Function

1. What is the purpose of classifying stationary points in scientific research?

The purpose of classifying stationary points is to identify and examine the critical points of a function, which can provide important information about the behavior and properties of the function. This can help scientists better understand the underlying mechanisms and relationships within their research.

2. How do you determine if a stationary point is a maximum or minimum?

To determine if a stationary point is a maximum or minimum, you can use the second derivative test. If the second derivative is positive, the stationary point is a minimum. If the second derivative is negative, the stationary point is a maximum. If the second derivative is zero, the test is inconclusive and further analysis may be needed.

3. Can a stationary point also be an inflection point?

Yes, a stationary point can also be an inflection point. An inflection point is a point where the concavity of a function changes. This can happen at a stationary point if the second derivative changes from positive to negative or vice versa.

4. What is the difference between a saddle point and a local extremum?

A saddle point is a stationary point where the function has a critical point but is neither a maximum nor minimum. It is a point of inflection where the slope of the function is zero, but the second derivative is neither positive nor negative. A local extremum, on the other hand, is a stationary point that is either a maximum or minimum of the function.

5. How can classifying stationary points be applied in real-world scenarios?

Classifying stationary points can be applied in various fields of science, such as physics, chemistry, and biology. For example, in physics, stationary points can help determine the equilibrium positions of objects in a system. In chemistry, they can aid in analyzing the stability and reactivity of molecules. In biology, they can be used to study the optimal conditions for growth and development of organisms.

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