Let $ \displaystyle L(x,y,z,\lambda) =(3x+2y+z+5)+\lambda(9x^2+4y-z)$
By using Lagrange Multiplier
$L_x=3+18\lambda x=0$
$L_y=2+4\lambda=0$
$L_z=1-\lambda=0$
It implies that $\lambda=1=-2$?
How should I do next?
Let $ \displaystyle L(x,y,z,\lambda) =(3x+2y+z+5)+\lambda(9x^2+4y-z)$
By using Lagrange Multiplier
$L_x=3+18\lambda x=0$
$L_y=2+4\lambda=0$
$L_z=1-\lambda=0$
It implies that $\lambda=1=-2$?
How should I do next?
Copyright © 2021 JogjaFile Inc.