0
$\begingroup$

How do I find the stationary points $(x,y,z)$ of this Lagrange problem? $$min \quad f(x,y,z)=5x^2-8xy-4xz+5y^2-4yz+8z^2+1$$ $$s.t.\quad x+y-4z=8$$

I know how to solve it using augmented matrix, but how do I solve it for $x, y$ and $z$?

For record, I usually know how to solve Lagrange and Kush-Kuhn-Tucker problems, but I struggled with this.

here is what I got

$f'x = 10x-8y-4z-\lambda=0$

$f'y =-8x+10y-4z-\lambda=0$

$f'z=-4x-4y+17z+4\lambda=0$

And I got $x=y$

1 Answers 1

3

If we set $g(x,y,z)=(x+y-4z)-8$, Lagrange's conditions give us that $$\nabla f(x,y,z) = \lambda\cdot \nabla g(x,y,z) $$ has to hold at every stationary point. That translates into: $$ \left\{\begin{array}{rcl}10x-8y-4z &=& \lambda\\ -8x+10y-4z&=&\lambda\\-4x-4y+16z &=& -4\lambda\end{array}\right.$$ where the third identity is a consequence of the first two identities.
It follows that the stationary points have the form $$ (x,y,z) = \left(\frac{\lambda}{2}+2z,\frac{\lambda}{2}+2z,z\right) $$ and in order to meet the constraint $g(x,y,z)=0$ we must have $\lambda=8$, or: $$ (x,y,z) = (4+2z,4+2z,z). $$ In such cases, we have: $$ f(x,y,z) = \color{red}{33}.$$

  • 0
    What do you mean by "third identity is a consequence of the first two identities"? – 2017-01-03
  • 0
    @Mataunited17: That $10x-8y-4z=\lambda$ and $-8x+10y-4z=\lambda$ imply $-4x-4y+16z=-4\lambda$. – 2017-01-03
  • 0
    A quick, but somehow related question. If I solve for one of the unknowns, and get $0=0$ when solving for a particular unknown, does it mean that the variable is free and has infinite solutions? – 2017-01-03