find the point on 2x+3y+z-11=0 for which 4x^2+y^2+z^2 is a minimum
Lagrangian is f−λg = 4x²+y²+z²−λ(2x+3y+z−11)
For optimality we need ( 8x, 2y, 2z ) = λ(2,3,1) → x=λ/4, y=3λ/2, z=λ/2
Sub in constraint equation gives λ(1/2+9/2+1/2) = 11 → λ=2
Hence only solution is ( 1/2, 3, 1 ) with f=11
At feasible point ( 1, 3, 0 ) f=13 so the point is a minimum
This could have been deduced from the fact that g is linear and f is positive definite
The answer is negative zero.
Comments
Lagrangian is f−λg = 4x²+y²+z²−λ(2x+3y+z−11)
For optimality we need ( 8x, 2y, 2z ) = λ(2,3,1) → x=λ/4, y=3λ/2, z=λ/2
Sub in constraint equation gives λ(1/2+9/2+1/2) = 11 → λ=2
Hence only solution is ( 1/2, 3, 1 ) with f=11
At feasible point ( 1, 3, 0 ) f=13 so the point is a minimum
This could have been deduced from the fact that g is linear and f is positive definite
The answer is negative zero.