Optimization, possibly just algebra help

[musicmar]

Homework Statement

Find the critical points of the function. Then use the second derivative test to determine whether they are local minima or maxima(or state that the test fails).

f(x,y)=(x-y)(e^(x2-y2))

The Attempt at a Solution

f_x=(x-y)(2x(e^(x2-y2)))+(e^(x2-y2))=0

f_y=(x-y)(-2y(e^(x2-y2)))-(e^(x2-y2))=0

(e^(x2-y2))((x-y)(2x)-1)=0

e^anything can never be 0, so:
2x²-2xy-1=0

f_y=(-2yx+2y-1)(e^(x2-y2))=0
(2y²-1)/(-2y)=x

Substituting into f_x:

2((2y²-1)/(-2y))²-2y((2y²-1)/(-2y))-1=0

((4y⁴-4y²+1)/-2y²)+2y²-2=0

-2y²+2-(1/2)y^-2-2=0

(-1/2)y^-2=0


Here is where I run into problems. y^-2 can never be 0.

If someone could check my derivatives and my algebra, that would be great.
Thanks.

 

[epenguin]

Well firstly the case that a derivative is never 0 would not be not a problem in itself - it would mean that the function has no extremum which is perfectly possible!

Your problem is at y=0.

Assuming no mistake, I think your result would sufficiently demonstrate that there is no extremum of the function except possibly at y=0. Where y=0 your treatment involving division by y which is valid everywhere else is invalid - you are not allowed to divide by 0!
If it is any consolation I believe Einstein once published a mistaken conclusion traced to having divided by something that was, not too obviously, 0.

In any case just look at the formulae for f_x and f_y (which you have not written out explicitly) and I think it can easily be seen that neither can be 0 at y=0.

Just put the two results f_x = 0, f_y = 0 side by side and it should be easy to eliminate between them without having to do any divisions. However there is more than one way to do this. If I am not mistaken (don't count on it) you can prove there is no extremum, and can do it without slogging (which is often a warning of a wrong or at least not best approach).

(Any equation you get eliminating between the two has to be true at an extremum - but the converse is not true!)
Slightly tricky.

 

[musicmar]

What do you mean by the "formulae for f_x and f_y"? f_x is simply the derivative of f with respect to x, is it not?

And by putting them side by side, do you mean I should set them equal to each other, because my professor specifically said not to do that.

 

[Dick]

What do you mean by the "formulae for f_x and f_y"? f_x is simply the derivative of f with respect to x, is it not?

And by putting them side by side, do you mean I should set them equal to each other, because my professor specifically said not to do that.

Write the two equations out without the exponential. x=0 doesn't lead to a critical point nor does y=0. So you can feel free to divide by y as you've done. I haven't checked all of the algebra, but I think you are correct to conclude that there are no critical points.

 

[epenguin]

Well you might have made a mistake, anyway I get
for the f_x factor

1 + 2x² - 2xy = g₁ say ...(1)

and for the f_y factor

-1 + 2y² - 2xy = g₂ say ...(2)Now actually from either of these in fact you can easily see that y=0 cannot possibly correspond to g₂ = 0 and therefore not to f_y=0 (nor, superfluously, to g₁=0 and f_x=0). So if you have already concluded that there are no critical points anywhere except possibly where y = 0 this now tells you there are no critical points anywhere.Concerning my other suggestion of elimination. At the critical point f_x=0 and f_y=0 which only happens as you say when g₁=0 and g₂=0.

So at the critical point

g₁ = g₂ = 0 .

So it is absolutely true that you can say that

g₁ = g₂ ...(3)

at a critical point! However without the =0 bit eq.(3) can also be true at points, maybe an infinite number of, that are not critical points. So you can see why your Prof. says do not do (3) because most students will just treat this as mindless equation grinding and just assume without thinking one equation they somehow got from another is true and so get false results.

But think also: if you show eq. (3) cannot be true anywhere, it does follow that there are no critical points!

Actually what I thought to do, as eq.(3) does not seem to lead to anything convenient was to say equally at the critical point

g₁ = -g₂ = 0

and just the first equation leads to x=y . But you can easily see from (1) and (2) that when x=y neither g₁ nor g₂ can be 0, so there cannot be any critical points.

Actually I think you can use for such a problem

Af_x = Bf_y

where A, B are anything (not 0) useful e.g. for eliminating one of x, y to obtain a critical point as long as you check the results. But as I do not remember having heard of such an approach it is open to criticism.

And my argument/calculation was simpler than yours but trickier so maybe your approach was safer!