Min max using partial derivatives

In summary, the function (x^2 + y^2)*e^(y^2 - x^2) has critical points at (0, 0), (1, 0), and (-1, 0) on the unit circle centered at (0, 0). The partial derivatives Fx and Fy are both equal to 0 at these points.
  • #1
StephenDoty
265
0
(x^2 + y^2)*e^(y^2 - x^2)
I am having trouble finding the critical points.

Fx=2xe^(y^2-x^2)(1-x^2-y^2)=0
Fy=2ye^(y^2 - x^2)(1+x^2 +y^2)=0

or
0=x(1-x^2-y^2)
0=y(1+x^2+y^2)

now, finding all the roots is giving me trouble. x and y obviously = 0, but I am unsure how to move forward to get the rest of the zeros, therefore giving me the critical points. Any help would be appreciated. Thanks.

Stephen
 
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  • #2
Fx = 0 iff x = 0 or x^2 + y^2 = 1
That is, Fx = 0 when x = 0 or when (x, y) is a point on the unit circle centered at (0, 0).

Fy = 0 iff y = 0 (1 + x^2 + y^2 >= 1 for all real (x, y) )
So both partials are 0 simultaneously for (0, 0), (1, 0), or (-1, 0).
 

1. What is the concept of min max using partial derivatives?

The concept of min max using partial derivatives involves finding the maximum or minimum value of a multivariable function by taking partial derivatives and setting them equal to zero.

2. Why is it useful to use partial derivatives in finding min max?

Using partial derivatives allows us to analyze a multivariable function by breaking it down into smaller parts, making it easier to find the maximum or minimum value. It also provides a systematic approach to solving optimization problems.

3. Can min max using partial derivatives be used for any type of function?

Yes, the concept of min max using partial derivatives can be applied to any type of function, including polynomial, exponential, logarithmic, and trigonometric functions.

4. How do partial derivatives help in determining whether a point is a maximum or minimum?

When taking partial derivatives, we are essentially finding the slope of the function in a particular direction. If the slope is positive, the point is a minimum, and if the slope is negative, the point is a maximum. If the slope is zero, further analysis is needed to determine the nature of the point.

5. Are there any limitations to using min max with partial derivatives?

One limitation is that the function must be continuous and differentiable over the domain of interest. Additionally, finding the partial derivatives and solving the resulting equations can be time-consuming for complex functions.

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