Finding local max, min and saddle points

In summary, the problem involves finding the critical points of the function f(x,y)=(1+xy)(x+y) by finding the partial derivatives and setting them equal to zero. After finding the first derivatives, the student encounters difficulty in finding the critical points and attempts various methods to solve for them, eventually realizing the importance of accounting for all possible values of x and y.
  • #1
maff is tuff
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1

Homework Statement



f(x,y)=(1+xy)(x+y)



Homework Equations





The Attempt at a Solution



I started out by expanding and got:

[itex]x+y+x^2y+xy^2[/itex]

Then I found all my partial derivatives and second derivatives:

[itex]f_{x}=1+2xy+y^2, f_{y}=1+2xy+x^2, f_{xx}=2y, f_{yy}=2x, f_{xy}=2(x+y)[/itex]

I know that both first partial derivatives must equal zero so I get:

[itex]f_{x}=1+2xy+y^2=0[/itex] and [itex]f_{y}=1+2xy+x^2=0[/itex]

This is the part I am stuck at; I can't find the critical points. I notice that there is symmetry so I tried subtracting the equations but I got y=x and got:

[itex]f_{x}=1+2x(x)+(x)^2=1+2x^2+x^2=0=1+3x^2=0---->x^2=-\frac{1}{3}[/itex]

I also tried setting [itex]f_{x}[/itex] and [itex]f_{y}[/itex] equal to each other but that didn't seem to work.

Thanks in advance for the help
 
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  • #2
I think I may have got it. When I got that [itex]x^2=y^2[/itex] I didn't account for that y could equal (-x). I did that and it got the right answer.
 

Related to Finding local max, min and saddle points

1. What is the process for finding local max, min, and saddle points?

The process for finding local max, min, and saddle points involves taking the partial derivatives of the function with respect to each variable and setting them equal to zero. Then, solving the resulting system of equations to find the critical points. Finally, using the second derivative test to determine the nature of each critical point and identify them as local maxima, minima, or saddle points.

2. What is the second derivative test used for in finding local max, min, and saddle points?

The second derivative test is used to determine the nature of critical points found by taking the partial derivatives and setting them equal to zero. It involves evaluating the second derivatives at each critical point. If the second derivative is positive, the critical point is a local minimum. If the second derivative is negative, the critical point is a local maximum. And if the second derivative is zero, further analysis is needed to determine the nature of the critical point.

3. How can I tell if a critical point is a saddle point?

A saddle point is a critical point where the second derivative is zero. To determine if a critical point is a saddle point, you can use the second derivative test. If the second derivative is zero, you must use other methods such as the first derivative test or graphing the function to determine the nature of the critical point.

4. Can a function have more than one local max or min?

Yes, a function can have more than one local max or min. This can happen when the function has multiple critical points, and the second derivative test shows that they are all local maxima or minima. In this case, the function will have multiple peaks or valleys within the given range.

5. Is it possible for a function to have no local max or min?

Yes, it is possible for a function to have no local max or min. This can happen when the function is either constantly increasing or constantly decreasing within the given range, meaning there are no critical points where the first derivative is equal to zero. In this case, the function will have neither a local maximum nor a local minimum.

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