Finding Max/Min Points of f(x,y) in R^2

  • Thread starter ArnfinnS
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In summary, the function f(x,y) = x^2*y*e^(-x^2 - 2y^2) for (x,y) in R^2 has two critical points at (1,0) and (-1/2,0). To classify these points, the second derivatives must be calculated and evaluated at each point.
  • #1
ArnfinnS
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hi...
i need to find the mex and min point of the function f(x,y) = x^2*y*e^(-x^2 - 2y^2) for (x,y) in R^2

here is what i tried : i found the partial derivatives :

f_x = 2x*y*e^8-x^2 - 2y^2) + x^2*y*e^(-x^2 -2y^2)*(-2x)

and f_y = x^2*y*(-4y)*e^(-x^2 - 2y^2)

i see that those partials equals 0 in the point (0,0). is this the only stationary point here?

what is max / and what is minimum?
can anyone help me?
 
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  • #2
I assume you mean the points where the plane tangent to the surface is parallel to the x-y plane, would this be where Fxy = 0? I've not quite got that far on my calc course.

Edit: oh by the way my graph below is a picture of what the general equation looks like,
 

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  • #3
ArnfinnS said:
(x,y) = x^2*y*e^(-x^2 - 2y^2) for (x,y) in R^2

Alright, this'll be a bit easier to read in Latex:

[tex]f(x,y)=x^2ye^{-x^2-2y^2}[/tex]


here is what i tried : i found the partial derivatives :

f_x = 2x*y*e^8-x^2 - 2y^2) + x^2*y*e^(-x^2 -2y^2)*(-2x)

and f_y = x^2*y*(-4y)*e^(-x^2 - 2y^2)

Alright, here's a first problem. You have:

[tex] f_x = 2xye^{-x^2 - 2y^2} - 2x^3ye^{-x^2 -2y^2}[/tex]
[tex]f_y = -4x^2y^2e^{-x^2 - 2y^2}[/tex]

The partial with respect to x looks fine, but you missed a term in the y partial. It should be:

[tex]f_y = x^2e^{-x^2 - 2y^2}-4x^2y^2e^{-x^2 - 2y^2}[/tex]

To find the critical points, you just set these to zero. This will eliminate those pesky exponentials.

[tex]-x^2+1=0[/tex]
[tex]-4y^2+1=0[/tex]

This will give you your critical points. I'll leave that part to you. Now, if you want to classify them, you have to calculate the second derivatives at your critical points and see if they're positive or negative in each direction. Do you know how to do that?
 
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Related to Finding Max/Min Points of f(x,y) in R^2

What is the definition of a local maximum/minimum point in a function?

A local maximum/minimum point in a function is a point where the function reaches its highest/lowest value in a specific region. This means that all points immediately around it have a lower/higher value.

How do you find the local maximum/minimum points of a function in two-dimensional space?

To find the local maximum/minimum points of a function in two-dimensional space, you can use the first and second derivative tests. The first derivative test involves finding the critical points of the function (where the first derivative equals 0 or is undefined) and then evaluating the second derivative at those points. If the second derivative is positive, the point is a local minimum. If the second derivative is negative, the point is a local maximum.

What is the difference between a local maximum/minimum point and a global maximum/minimum point?

A local maximum/minimum point is a point where the function reaches its highest/lowest value in a specific region, while a global maximum/minimum point is the highest/lowest value of the entire function. In other words, a global maximum/minimum point is the largest/smallest value of the entire function, while a local maximum/minimum point is the largest/smallest value in a specific region.

Can a function have more than one local maximum/minimum point?

Yes, a function can have more than one local maximum/minimum point. This can occur when the function has multiple peaks/valleys or when there are multiple local maximum/minimum points in a specific region.

What is the significance of finding the local maximum/minimum points of a function?

Finding the local maximum/minimum points of a function can help us understand the behavior of the function and identify important features such as peaks, valleys, and points of inflection. This information can also be used to optimize the function for various applications.

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