Question

Can anyone help me with Lagrange Multipliers? find the point in the first quadrant on the curve xy^4=31 that is closest to the origin.

Answer 1

use x^2+y^2+z^2 -a^2
for your distance function
grad (x^2+y^2+z^2 -a^2) = lambda *grad(xy^4-31)

find the lambdas that give positive x and y

Answer 2

where is the z^2-a^2 coming from?

Answer 3

it's a general form for 3 space. you'll use it for any function g(x,y,z). here you notice that it won't matter because there's no k coordinate of the gradient of your curve, so 2z=0

Answer 4

alright so i did that and i got the 3 equations but i get stuck. i was able to do a problem simmilar but there was not both x and y in a partial.

Answer 5

3?

Answer 6

counting 2z=0 ?

Answer 7

umm no ill type out what i have one moment

Answer 8

2x = lambda(y^4)
2y= lambda(4xy^3)

Answer 9

\[2x=\lambda(xy^4-31)\]
\[2y=\lambda(y^4-31)\]
\[xy^4=31\]

Answer 10

whoops your right i dident mean to add the 31's in any of those

Answer 11

what's the gradient of xy^4-31

Answer 12

<y^4,4xy^3> i really messed up typing those lol sorry

Answer 13

+ 0 k yes, right.

Answer 14

so using those equations should i love for lambda first?or x and y?

Answer 15

solve the second for y^2 and sub.s into the first.

Answer 16

so y^2=4((lambda)^2)(x^6)?

Answer 17

you don't have to do it that way, that's just what I'd do..

Answer 18

y^2 = 1/(lambda*2x)

Answer 19

ehhh i dont understand what im trying to get to. in the end i need lambda, x, and y right?

Answer 20

lambda, really...

Answer 21

if you find what values lambda must have it tells you how x and y must be related for your constraint to hold.

Answer 22

alright i found x=lambda(y^4)/2 and y=\[\sqrt[8]{62/\lambda}\]

Answer 23

but for some reason i feel like im making this harder then it needs to be

Answer 24

looks like you're going to have to sub.s back into the original equation to find lambda for this problem..

Answer 25

ok i thought so i started to do that and i got
\[(x^2\lambda2x+1)/\lambda2x\]