Lagrange multipliers on the unit sphere

Question

I have a function $f(x,y,z)=x^2+xy+y^2+yz+z^2$ which I'm supposed to maximize on the Unitsphere $B_1(0)=\{x \in \mathbb{R^3}:\Vert x\Vert_2 \le 1\}$ and I have a few questions concerning the process of the computation:

1. My first question is if I'm allowed to transform the condition into $\{x \in \mathbb{R^3}:\Vert x\Vert_2^2 = 1\}$ so that it's $g(x,y,z)=x^2+y^2+z^2=1 \quad$ I'd assume this would work since the function grows larger for growing $x,y,z$, so the Maximum under my condition will be on the edge of the sphere. Furthermore since it is a unit-vector the norm is equal to the square of the norm.
2. If this is possible/correct I have some trouble with the algebraic part of the computation. When setting up my functions I'm left with: $$\begin{align}&I. 2x+y=\lambda2x \\& II. 2y+x+z=\lambda2y \\& III. 2z+y=\lambda2z \\& IV. x^2+y^2+z^2=1 \end{align}$$

Whatever I do, I can't seem to rewrite the equations in such a way that this problem is solvable. The furthest I can show is that $x=z$. Any further rewriting I do yields me something which is dependent on $\lambda$ and one of the other three variables. What am I missing? Thanks in advance.

Answer 1

The first three equations are equivalent to$$\left\{\begin{array}{l}(2-2\lambda)x+y=0\\x+(2-2\lambda)y+z=0\\y+(2-2\lambda)z=0,\end{array}\right.\tag1$$which is a system of three linear equations in the unknowns $x$, $y$, and $z$, depending upon a parameter $\lambda$. The matrix of this system is$$\begin{bmatrix}2-2\lambda&1&0\\1&2-2\lambda&1\\0&1&2-2\lambda\end{bmatrix}.\tag2$$The system has a solution other than $(0,0,0)$ if and only if the determinant of $(2)$ is $0$. It is cleat that that occurs if $\lambda=1$. On the other hand, the determinant of $(2)$ is $-4(2\lambda^3-6\lambda^2+5\lambda-1)$ and we already know that $1$ is a root of this polynomial. Using this, it is easy to see that the other roots are $1\pm\frac1{\sqrt2}$. So, for each $\lambda\in\left\{1,1+\frac1{\sqrt2},1-\frac1{\sqrt2}\right\}$, there are vectors $(x,y,z)\in\Bbb R^3\setminus\{(0,0,0)\}$ such that $(x,y,z)$ is a solution of $(1)$. Computes them, divide them by their norms, and you will get the six solutions of your original system. For each of them, compute $f(x,y,z)$ and see where is that you get the maximal value.

Answer 2

$2x+y = 2\lambda x$

$2y+x+z = 2\lambda y$

$2z+y = 2\lambda z$

$x^2+y^2+z^2=1$

From first equation $\lambda = \frac{2x+y}{2x}$

Now plugging it into $2$nd and $3$rd equations, we have $3$ equations in $3$ variables.

The solution comes to $x = z = \pm \frac{1}{2}, y = \pm \frac{1}{\sqrt 2}$. (I took help of Wolfram Alpha)

On your first point, given the function $xy + yz + x^2 + y^2 + z^2, $ you can absolutely argue that the maximum will be on the sphere and not inside. You see why?

Last but not least, while you need to maximize $f(x, y, z) = xy + yz + x^2 + y^2 + z^2$, as we know the maximum value of $x^2 + y^2 + z^2 = 1 \,$, I suggest you try and maximize the function $g(x, y, z) = xy + yz$. You get the same result as I mentioned above but it makes things much simpler to solve by hand.