Find any local max or min of $x^2+y^2+z^2$ s.t $x+y+z=1$ and $3x+y+z=5$

Question

Find any local max or min of \begin{align} f(x,y,z)=x^2+y^2+z^2 && (1) \end{align} such that \begin{align} x+y+z=1 && (2)\\ 3x+y+z=5 && (3) \end{align}

My attempt. Let $L(x,y,z,\lambda_1, \lambda_2)= f(x,y,z)+\lambda_2 (x+y+z-1) + \lambda_1 (3x+y+z-5)$

$L_x=2x+ 3 \lambda_1 + \lambda_2 =0$

$L_y=2y+ \lambda_1 + \lambda_2=0$

$L_z=2z+\lambda_1 + \lambda_2=0$

Solve for $x,y,z$ we get:

$x=\frac{-3 \lambda_1 - \lambda_2}{2}$

$z=y=\frac{-\lambda_1 - \lambda_2}{2}$

with the use of $(2)$ and $(3)$ $\implies$

$x=2$

$y=z= \frac{-1}{2}$

so the stationary point is $(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$

The Hessian of $L$ gives a postive definite matrix for all $(x,y,z)$ thus

$(x,y,z)=(2, \frac{-1}{2},\frac{-1}{2})$ is the only local minimizer of $f$ and there is no maximizors of $f$.

Is this correct?

Answer 1

Your result is correct. This is an alternative approach.

Here we have two non-parallel planes which intersect along the line $(x,y,z)=(2,t,-1-t)$ where $t\in \mathbb{R}$. Hence we reduce the problem to a $1$-variable case, $$f(x,y,z)=4+t^2+(-1-t)^2=2t^2+2t+5$$ The above quadratic polynomial has just one local/global minimum at $t=-1/2$ that is at $(2,-1/2,-1/2)$. There is no local/global maximum.

Answer 2

An option:

1) $x+y+z=1$; and

2) $3x+y+z=5$;

$2$ planes , their intersection is a straight line.

Subtract: 2)-1):

$2x=4$; $x=2$ ;and

$y+z=-1$;

$d^2=x^2+y^2+z^2$ .

Minimal distance of line from origin:

$d^2= 4 +y^2+z^2.$

2D problem:

Minimize $y^2+z^2$ with constraint $y+z=-1$.

$d_2^2= $

$[-(1+z)]^2+z^2=2z^2+2z+1=$

$2(z^2+z)+1= $

$2[(z+1/2)^2]-1/2+1\ge 1/2$.

Equality at $z=-1/2$;

Finally:

Minimum at :

$x=2$ ; $y=-1/2$; $z=-1/2$;

$d^2_{\min}= 4+1/2=9/2;$