# Lagrange multipliers and two constraints

by autre
Tags: constraints, lagrange, multipliers
 P: 117 So I need to find the min and max values of f(x,y,z) = x^2 + 2y^2 + 3z^2 given the constraints x + y + z = 1 and x - y + 2z =2. I've gotten as far as (2x, 4y, 6z) = (u,u,u) + (m,-m,2m). I'm stuck trying to solve this system of equations. Any hints?
 HW Helper Thanks P: 10,769 You have the constraints as two more equations: x+y+z=1 and x-y+2z=2. ehild
HW Helper
Thanks
P: 5,200
 Quote by autre So I need to find the min and max values of f(x,y,z) = x^2 + 2y^2 + 3z^2 given the constraints x + y + z = 1 and x - y + 2z =2. I've gotten as far as (2x, 4y, 6z) = (u,u,u) + (m,-m,2m). I'm stuck trying to solve this system of equations. Any hints?
Write them out: 2x = u+m, 4y = u-m, 6z = u+2m. Solving for x, y and z as functions of u and m is quite easy. Now you need two more equations to determine u and m. Can you guess what they are?

RGV

 HW Helper P: 3,307 Lagrange multipliers and two constraints whilst lagrange is a good idea, and the comments above from ehild & Ray are good ideas about where to head - i think a geometric method would be a tiny bit quicker here... though its upto preference i guess, and all the same at the end of the day the intersection of two planes is a line. If you find the equation of that line it reduces to optimising a single variable function.
 P: 1 how would you go about solving for x,y,z as functions of u and m?