Question

Lagrange Multiplier question -

How does one solve for their lamba (or even just get the x,y,z etc straight away) when the constraint has higher powers and variables multiplying each other?

Example - x^2 + y^2 subject to 17x^2 + 12 xy + 8y^2 = 100

When solving, I end up with a non-linear system, and don't know how to go on.

Answer 1

Let f(x,y)=x^2 + y^2 , and g(x,y)= 17x^2 + 12 xy + 8y^2 - 100
Then df(x,y)+\(\lambda\)dg(x,y)=0
which gives you together with g(x,y)=0 system of 3 equations for 3 unknowns:
2x+\(\lambda\)34x+12y=0
2y+\(\lambda\)12x+16y=0
17x^2 + 12 xy + 8y^2 - 100 =0

Answer 2

3equations and 4unknows^

Answer 3

which is the fourth?

Answer 4

is the answer lamda = 1/17, -1/18 ? if thats right, i can explain how i got the answer

Answer 5

Thanks, I have that much sorted already though.

From there I go to λ = x/(17x+6y) = y/(6x - 8y), and can't get them on their own.

Answer 6

ow oops, sorry myko, my bad... didnt look at that properly

Answer 7

i hope i did that right^

Answer 8

you didn't take differentials, :)

Answer 9

http://www.bilkent.edu.tr/~sertoz/courses/math114/2011/mid2-sol.pdf
page 4, Q3

Answer 10

Why is your partial derivative restricted to a singe term. Do the entire expression.

Answer 11

My bad. Should be:
2x+λ (34x+12y)=0
2y+λ (12x+16y)=0
17x^2 + 12 xy + 8y^2 - 100 =0

Answer 12

Thanks guys, I have it now, I was just trying to solve for lambda at the wrong place. I started by getting y in terms of only lambda and x, then then subbing that so I have x in terms of lambda, and then it's simple.