Lagrange multipliers f(x,y) = x^2 + y^2 - x -y +1

[xoninhas]

find the maximum extreme, using lagrange multipliers method, for f(x,y) = x^2 + y^2 - x -y +1 in a disc of radius 1 and center in origin (0,0).

My result was two extremes = (?(2)/2, ?(2)/2) and (-?(2)/2, -?(2)/2) which the maximum was the later with ?(2) + 2

Is my result right or you guys got something else? I graphed the function and it doesn't seem right... (i used grapher on mac os)

 

[tkhunny]

and it doesn't seem right...

This statement should be a warning sign to your brain. It suggests that you are lost and wandering. Mathematics is never about "seeming". Prove it or don't prove it.

You need to rethink the problem just a little, but only because of one word. Did you really mean "Disc", or did you mean "Circle"?

If you mean "Circle", which you did not state,then you're good.
If you mean "Disc", then you may wish to investigate (1/2,1/2).

 

[xoninhas]

In fact it is disc... but why should I investigate (1/2,1/2) where did that come from?? how do I get to that? I think I did everything like it is supposed to be done. It does say JUST using lagrange multipliers method... with the f'(x,y) = 0 method I also get (1/2, 1/2) but am I suppose to use it?

first I get

gradiet = G lambda = L

Gf(x,y) = LGF(x,y)
F(x,y) = 0

which I got

2x - 1 = 2xL
2y - 1 = 2yL
x^2 + y^2 = 1

then:

2x(1 - L) =1
2y(1 - L) =1

from what I got:
x=y
2*x^2 = 1 => x = ±(?2)/2

Where should have I seen the (1/2, 1/2)

 

[tkhunny]

It could be a question problem. How many times have I stated the magnificent difficulty in writing a clear and complete problem statement?

The Lagrange Multiplier imposes the constraint "=1". This is the circle, not the disk. The disk would be "<=1". Can you think of a way to impose this more general constraint?

 

[xoninhas]

Ok, I'll transmit that to my teacher... because IN FACT this is from a test, and it is correctly translated trust me. I thought there was something weird, so it is in fact in the question... I'm already not the best student at maths, but when even the questions are wrong it's even harder don't you agree?

By the way for the point (1/2, 1/2) it's a local minimum right? and the other two got by lagrange method are absolut maximum and minimum. Right?

I'm sorry if you feel you lost your time in vain, but you didn't! Thanks once again!! This lagrange thing is actually quite simpler than I thought!

EDIT:
As for imposing the more general condition, no I have no idea how to do it so to use lagrange...

 

[tkhunny]

Very good. i can hardly wait for the repsonse.

f(1/2,1/2) is a Global Minimum.

The vlocations the Lagrange Multiplier method gave you are Constrained Extrema. They are NOT Global Extrema. I suppose your textbook or teacher could disagree with me on this, as it may be that no one ever calls things "Constrained Extrema".