09-03-2005 00:56									#1
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Im doing Lagrange multipliers and i cant seem to figure out which of the possible solution are correct. heres the problem.

f(x,y) = x^2+y^2 (maximize/minimize)
g(x,y) = 4x^2+Y^2=1 (constraint)

find closest and furthest pts from the origin

fx(x,y) = 2x
fy (x,y) = 2y
gx(x,y) = 8x
gy(x,y) = 2y

∇f (x,y) = λ∇g(x,y)

∇f (x,y) = <2x, 2y>
∇g(x,y) = <8x, 2y>

<2x, 2y> = λ<8x, 2y> ; 4x^2+y^2=1

2x=λ8x
2y=λ2y
4x^2+y^2=1

x=λ4x
y=λy
4x^2+y^2=1

x-λ4x=0
y-λy=0
4x^2+y^2=1

roots:
x(1-4λ)=0 . . . y(1-λ)=0
x=0 or λ=1/4 . . . y=0 or λ=1

so ive found the roots, now i need to find the possible solutions. how is that done? do you just match all of them together like so?

1. x=0 y=0
2. y=0 λ=1/4
3. x=0 λ=1
4. x=0 λ=1/4
5. y=0 λ=1

or do some of them not make it to the possible solutions part? and from here, how do you choose the correct selections (2 of them right?) to plug into the contraint (4x^2+y^2=1), where we find the roots, and then plug iinto the same eqn again to fnd the solution.

the final solution is

f(0,1) = 1
f(1/2,0) = 1/4

the closest pts to the origin (1/2,0) and (-1/2,0)
the farthest pts to the origin (0,1) and (0,-1)

thanks.

	09-03-2005 01:43									#2
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It's easier to solve this without Lagrange multipliers. Just find y(x) and than you can find df/dx .

	09-03-2005 02:11									#3
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quote: x-λ4x=0 y-λy=0 4x^2+y^2=1

For a non-trivial LaGrange multiplier, the first two equations imply that either x = 0 or y = 0.

If x = 0, the constraint equation implies that y = +/- 1.

If y = 0, the constraint equation implies that x = +/- 1/2.

So the extrema are (0, +/- 1) and (+/- 1/2, 0). The former coordinates are obviously the maxima (f = 1) and the latter the minima (f = 1/4) for f.

	09-03-2005 03:47									#4
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Ramo, one day I hope you come into a Math Poly thread and just post "asd hahahaha I don't care anymore, I is happy drunk respect!!! hit y'alll!"

	09-03-2005 04:01									#5
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Yeah, but I'm only at poly when at work. And I don't get to work drunk .

	09-03-2005 04:38									#6
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quote: It's easier to solve this without Lagrange multipliers. Just find y(x) and than you can find df/dx .

i must use the multipplier.

quote: For a non-trivial LaGrange multiplier, the first two equations imply that either x = 0 or y = 0. If x = 0, the constraint equation implies that y = +/- 1. If y = 0, the constraint equation implies that x = +/- 1/2. So the extrema are (0, +/- 1) and (+/- 1/2, 0). The former coordinates are obviously the maxima (f = 1) and the latter the minima (f = 1/4) for f.

oh thats really easy. so we dont need to do anything with the multiplier then. we just find what x and y are equal to, plug x into the constraint, plug y into the constraint and solve.

so wahts the point of lambda then?

	09-03-2005 04:58									#7
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You only got the equations through using the multiplier.