Find infimum and supremum of the function $f$ defined by the formula $f(x, y, z) = 2x+ 2y-3z$ on the set $A$

Question

Let $A = \{(x, y, z) \in \mathbb R^3 : x^2 + y^2 + z^2 = 2, xy + yz +zx +1 = 0\}$. Find infimum and supremum of the function $f$defined by the formula $f(x, y, z) = 2x+ 2y-3z$ on the set $A$.

I want to use Lagrange multipliers so I got the following system of equations: $$ \begin{cases} 2= \alpha x + \beta y + \beta z \\ 2= \alpha y + \beta x + \beta z \\ -3 = \alpha z+ \beta x + \beta y \\ x^2 + y^2 + z^2 =2 \\ xy + yz + zx +1 = 0 \end{cases} $$

Then: $$ \begin{cases} 2= \alpha x + \beta y + \beta z \\ 2= \alpha y + \beta x + \beta z \end{cases} \Rightarrow \alpha (x - y)+\beta(y-x)=0 \Rightarrow (x-y)(\alpha - \beta) =0 \Rightarrow y=x \text{ or } \alpha = \beta $$

First case:

5. equation: $$x^2+2xz+1=0 \Rightarrow 2xz=-1-x^2 \Rightarrow z = \frac{-1-x^2}{2x}$$

6. equation: $$2x^2+\frac{1+2x^2+x^4}{4x^2}=2 \Rightarrow 9x^4+2x^2+1-8x^2=0 \Rightarrow 9x^4-6x^2+1=0 \Rightarrow x = \pm \frac{1}{\sqrt 3}$$

Coordinates found: $(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},-\frac{2\sqrt 3}{3})$, $(-\frac{1}{\sqrt 3}, -\frac{1}{\sqrt 3},\frac{2\sqrt 3}{3})$.

But then I get contradiction. For example for $(\frac{1}{\sqrt 3},\frac{1}{\sqrt 3},-\frac{2\sqrt 3}{3})$: $\frac 92 = \sqrt{3} (\alpha - \beta)$ and $6=\sqrt 3 (\alpha - \beta)$

Second case:

$ 2 = \alpha (x+y+z)$ and $-3=\alpha (x+y+z)$ contradiction

Could someone check where I made a mistake? I don't think this problem has no solution...

Answer 1

Consider

$ (x + y + z)^2 = x^2 + y^2 + z^2 + 2 (x y + x z + y z ) = 2 + 2(-1) = 0 $

Therefore, the intersection of the two constraints is the same as the intersection of $x^2 + y^2 + z^2 = 2 $ and $x + y + z = 0 $.

The intersection of the sphere and the plane is a circle of radius $\sqrt{2}$ and is spanned by two vectors that are orthogonal to the normal vector of the plane.

We can find two orthogonal unit vectors $u_1, u_2$ that are orthogonal to the normal of $x+y+z=0$ which is $[1, 1, 1]$, and these unit vectors are, for example,

$ u_1 = [1, -1, 0] /\sqrt{2} $

$ u_2 = [1, 1, -2] / \sqrt{6} $

Hence,

$(x, y, z) = \sqrt{2} \left( \cos t \ u_1 + \sin t \ u_2 \right) $

$ f = 2 x + 2 y - 3 z = [2, 2, -3] \cdot \sqrt{2} \left(\cos t \ u_1 + \sin t \ u_2 \right) \\ = \sqrt{2} ( [2, 2, -3] \cdot u_1 \ \cos t + [2,2,-3] \cdot u_2 \ \sin t ) $

$ f = \sqrt{2} ( \dfrac{10}{ \sqrt{6} } \sin t ) $

Therefore, the maximum of $f$ is $\dfrac{10}{\sqrt{3} } $ and it occurs at $t = \dfrac{\pi}{2} $ at which, we have

$ (x, y, z) = \dfrac{1}{\sqrt{3}} [ 1, 1, -2 ] $

and the minimum is $-\dfrac{10}{ \sqrt{3}}$ and occurs at $ t = \dfrac{3\pi}{2} $, at which, we have

$(x, y, z) = - \dfrac{1}{\sqrt{3}} [1, 1, -2] $

Answer 2

The problem cannot be solved directly using the method of Lagrange multipliers. In fact, consider the map $$ \begin{array}{rccc} g:&\mathbb{R}^3&\longrightarrow&\mathbb{R}^2\\ &(x,y,z)&\mapsto&\left(x^2+y^2+z^2,xy+yz+zx\right). \end{array} $$ Let $g_1$ and $g_2$ be the first and the second component of $g$ respectively. Your problem is to find the maximum and the minimum of the restriction of $f$ to $g^{-1}\{(2,-1)\}$. You tried to solve the system $$ \left\{\begin{array}{l} \nabla f(x,y,z)=\alpha\nabla g_1(x,y,z)+\beta\nabla g_2(x,y,z)\\ g(x,y,z)=(2,-1) \end{array}\right. $$ and you got no solutions. That's natural, because there is an assumption missing: the method works as long as $g'$ is a surjective map. And, when $(a,b,c)\in g^{-1}(\{(2,-1)\})$, $g'(a,b,c)$ is not surjective. In order to see why, note that $g'(a,b,c)$ is the map from $\mathbb{R}^3$ into $\mathbb{R}^2$ defined by $$ (x,y,z)\mapsto(2ax+2by+2cz,(b+c)x+(a+c)y+(a+b)z).\label{sys}\tag1 $$ On the other hand \begin{align} (a+b+c)^2&=a^2+b^2+c^2+2(ab+bc+ca)\\ &=2-2\\ &=0, \end{align} and therefore $a+b+c=0$. So, $c=-a-b$ and then the map \eqref{sys} becomes $$ (x,y,z)\mapsto(2ax+2by-2(a+b)z,-ax-by+(a+b)z), $$ which is not surjective since for each point $(\alpha,\beta)$ in its range you have $\alpha=-2\beta$.

In order to use the method of Lagrange multipliers, you can change the problem a little. It follows from the computations made above that, if $(a,b,c)\in\mathbb{R}^3$ is such that $a^2+b^2+c^2=2$, then $ab+bc+ca=-1\iff a+b+c=0$. So, apply the method of Lagrange multipliers, but this time using the function $$ \begin{array}{rccc} h:&\mathbb{R}^3&\longrightarrow&\mathbb{R}^2\\ &(x,y,z)&\mapsto&\left(x^2+y^2+z^2,x+y+z\right) \end{array} $$ and the point $(2,0)$. The map $h'$ is surjective at every point $(a,b,c)\ne(0,0,0)$ and the method will then give you the correct answer.