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March 19th, 2012, 11:10 AM   #1
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Lagrange multiplier (am I doing this right?)?

Hi guy's, could someone please tell me if I'm doing this right!?

the question that was asked:

f (x, y) = 18y + 6x ? 2xy

Calculate (by Lagrange multiplier) the stationary points of f(x,y), under the constraint:

x ? y = 0

Also calculate the value of the function at the stationary points and also calculate 'Lagrange multiplier' (?) for the(se) points.

So what i did:

L(x,y,?)= 18y+6x-2xy-?(x-y)

L'x = 6-2y-?=0
L'y = 18-2x-2?y=0

constraint: x-y=0
This means: x=y
y= ?x or y=-?x

18-2x-2?y = 0

18-2y-2?y=0

?=(18-2y)/2y = (6-2y)/1

18-2y = 12y-4y

18= 12y-2y

This gives me:

-2y+12y-18=0

On this is used the ABC formula which gave me: X(1)= 3 and X(2)= 3. Meaning x = 3

So y=?3 or y=-?3

Corresponding points (3,?3) (3,-?3)

I would really really appreciate it if someone could check this for me!!
Thank you for your attention

p.s. sorry if its a bit messy, and my apologies if i have posted this in the wrong section.
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March 19th, 2012, 12:05 PM   #2
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Re: Lagrange multiplier (am I doing this right?)?

To find the extrema of subject to the constraint , solve the system of equations:







In our case, we have:

and

so we solve the system:







From the first 2 equations, we see:





Substituting for x into the 3rd equation, we get:









i) When we have:

and and

This is the constrained minimum.

ii) When we have:

and and

This is the constrained maximum.
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March 25th, 2012, 07:14 AM   #3
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Re: Lagrange multiplier (am I doing this right?)?

Thank you soooooo much Mark!! Cant thank you enough
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