Question

I have a question on Lagrange Multipliers: if f(x,y)=e^(-xy) and the constraint is x^2+4y^2-1<=0, how do I find the critical points when the case is x^2+4y^2-1<0?

Answer 1

i.e. \[x^2+4y^2-1<0\]

I've already found the critical points for when \[x^2+4y^2-1=0\]

And I know that df/dx=-ye^(-xy) and df/dy=(-xe^(-xy))

Answer 2

do your normal equality constraint stuff on the ellipse, which seems you have done

and then look for a CP, ie \(f_x = f_y = 0\), within the ellipse.

you could luck out if, say, there was a local max or a min in there.... then you know that's what you are looking for, optimisation-wise.

but here, **think** you'll find a saddle point at (0,0), eg note \(f(x,y) = 1\) all along the x and y axes as well as at origin......and you have some kind of symmetry going on as the exponent is \(-xy\) which will be -ve in Q1, Q3 and vice versa in others

i'm getting \(e^{1/4}\) (1.28) and \(e^{-1/4}\) (0.78) on the ellipse itself ie along \(y = \pm {1 \over 2} x\)

Answer 3

picture paints a thousand words

https://www4b.wolframalpha.com/Calculate/MSP/MSP35641h4d9da385e6bf980000259aa58g8078ic3g?MSPStoreType=image/gif&s=57

Answer 4

@IrishBoy123
I get four solutions that satisfy the Lagrange conditions,
\(x=\pm1/\sqrt 2, y=\pm x/2,\lambda=e^{1/4}/4\) and
\(x=\pm1/\sqrt 2, y=\mp x/2,\lambda=-e^{-1/4}/4\)
all of which are for equality (=0) in the constraint.
I'm not sure how to take care of the <0 case(s). Any ideas?

Answer 5

@mathmate

first of all, there is a way to solve Lagrange Multiple type optimisations with inequalities in the constraints. It involves the use of **slack variables**. i find it absolutely horrible. It seems to me to undo the point of the method, which is simplified algebra, by adding more variables. i don't know if you know that approach, i have never finished one off.

so inside the constraint, i see no problem in just looking for fixed points. if you find a max or min inside the ellipse, then that is a local max or min for the ellipse and its region. job done.

here, there is no max or min. \(f_x = f_y = 0\) occurs at the origin. but it is a saddle. so its not a max or a min so forget about it!

i can't remember if i looked at the Hessian when i first did this but what you can do here, because the function is so clean, is you can draw contour lines. so the line \(y = 1/x\) has \(f(x,y) = 1/e\), the line \(y = 2/x\) has \(f(x,y) = 1/e^2\) etc . so \(f(0,0) = 1\) and the function grows exponentially in Q2 & Q4 and dies exponentially in Q1 & Q3 . sure you know all that, just saying.

you can then see how the saddle is symmetric and in particular, because it is not a max or min, and so can be ignored.

not sure if that helps.

btw, i agree with location of your fixed points on the ellipse but i don't get the 4 denominator when i plug those back into \( f(x,y) \) to get actual values.

Answer 6

@irishboy123
Yes, it is a saddle point. I had it plotted and it's quite visual.
http://prnt.sc/c2pxb2

Even though (0,0) satisfies constraints, it would not be the point we're looking for.
The four points I gave when substituted back into f(x,y) all gave e^(-1/4) (approx. 0.7788), which are less that f(0,0)=1.

As a check, I solved for x^2+4y^2=1/2 and =1/4, both of which gave values between 0.7788 and 1, probably evident from the fact that the constraint is an ellipse, so f(x,y) along y=\(\pm\)x/2 is expected to be monotonic.

It's unfortunate that Lagrange multipliers handles inequality constraints with difficulty.
Yes, I understand the role of slack variables from linear programming. It might be worth a try for me, some time down the road!

Thank you for the detailed explanations.

Answer 7

brilliant! thank you.

Answer 8

wwow

Answer 9

@mathmate

i just remembered something else as well - The Extreme Value Theorem is the proper name given to this idea, of just looking for CP's in the region

I'm sure you know that already. but i thought i'd mention.

Answer 10

@irishboy123
Thank you, that's a good reminder. Sometimes we do things by reflex. lol

Answer 11

...

Answer 12

@IrishBoy123 @mathmate Hey guys! I'm so sorry I just got back to this now. Thank you so much, that helps out a lot.

Answer 13

first of all, there is a way to solve Lagrange Multiple type optimisations with inequalities in the constraints. It involves the use of **slack variables**. i find it absolutely horrible. It seems to me to undo the point of the method, which is simplified algebra, by adding more variables. i don't know if you know that approach, i have never finished one off.

so inside the constraint, i see no problem in just looking for fixed points. if you find a max or min inside the ellipse, then that is a local max or min for the ellipse and its region. job done.

here, there is no max or min. fx=fy=0 occurs at the origin. but it is a saddle. so its not a max or a min so forget about it!

i can't remember if i looked at the Hessian when i first did this but what you can do here, because the function is so clean, is you can draw contour lines. so the line y=1/x has f(x,y)=1/e, the line y=2/x has f(x,y)=1/e2 etc . so f(0,0)=1 and the function grows exponentially in Q2 & Q4 and dies exponentially in Q1 & Q3 . sure you know all that, just saying.

you can then see how the saddle is symmetric and in particular, because it is not a max or min, and so can be ignored.

not sure if that helps.

btw, i agree with location of your fixed points on the ellipse but i don't get the 4 denominator when i plug those back into f(x,y) to get actual values.

Answer 14

@JSnake789
Yes, your comments pretty much confirm what @irishboy123 expressed earlier that the use of slack variables is counter-nature for inequality constraints.

Well, all four solutions give f(x,y)=\(e^{-1/4}\) or 0.7788. The value of \(\lambda\) is not used in calculating f(x,y), just for the optimization part. Or perhaps I misunderstood your comment about the denominators.

Answer 15

@satellite73 can you warn me? plzzz

Answer 16

:3

Answer 17

:3

Answer 18

.-.

Answer 19

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Answer 20

I have a really easy question!!

Answer 21

Would one of y'all help me?
Given the equation, y = 3(0.87)^x , where y is the amount of money in thousands in x years, describe what is happening?
The amount of money is decreasing at a rate of 13% per year
The amount of money is decreasing at a rate of 87% per year
The amount of money is decreasing at a rate of 13 per year
The amount of money is decreasing at a rate of 87 per year

Answer 22

@redheadangel
Please post a new question, since this question does not relate to the original post.

Answer 23

...??