Minimization on compact region

Question

I need to solve the minimization problem

$$\begin{matrix} \min & x^2 + 2y^2 + 3z^2 \\ subject\;to & x^2 + y^2 + z^2 =1\\ \; & x+y+z=0 \end{matrix}$$

I was trying to verify the first order conditions using lagrange multipliers

$$\left\{\begin{matrix} 2x = 2\lambda_1x + \lambda_2 \\ 4y = 2 \lambda_1 y + \lambda_2 \\ 6z = 2\lambda_1 z + \lambda_2 \end{matrix}\right.$$

Summing the 3 equations we get $$ 2y + 4z = 3 \lambda_2$$ Multiplying the first one for x, the second for $y$ and third for $z$ and summing: $$ 2 + 2y^2 + 4z^2 = 2 \lambda_1$$

I just got stuck there and I have no idea how to proceed. We can check the answer here but I couldn't find none of the variables. Any idea?

Thanks in advance!

Answer 1

since you have $$x=\dfrac{\lambda_{2}}{2(1-\lambda_{1})},y=\dfrac{\lambda_{2}}{4-2\lambda_{1}},z=\dfrac{\lambda_{2}}{6-2\lambda_{1}}$$ and $x+y+z=0$ so we have $$\dfrac{\lambda_{2}}{2}\left(\dfrac{1}{1-\lambda_{1}}+\dfrac{1}{2-\lambda_{1}}+\dfrac{1}{3-\lambda_{1}}\right)=0$$ so we have$\lambda_{2}=0$ or $$\dfrac{(2-\lambda_{1})(3-\lambda_{1})+(1-\lambda_{1})(3-\lambda_{1})+(1-\lambda_{1})(2-\lambda_{1})}{(1-\lambda_{1})(2-\lambda_{2})(3-\lambda_{3})}=0$$ then we have $$3\lambda^2_{1}-12\lambda_{1}+11=0\Longrightarrow \lambda_{1}=2\pm\dfrac{1}{\sqrt{3}}$$ then you can easy to find this respect $\lambda_{2}$

Answer 2

Here is how I would solve this problem using Gröbner bases in Macaulay2.

R = QQ[x,y,z,l1,l2] -- define a ring

I = ideal (2*x - 2*l1*x - l2, 4*y - 2*l1*y - l2, 
           6*z - 2*l1*z -l2, x^2+y^2+z^2-1, x+y+z) 
-- System of polynomial equations has isolated solutions:

i7 : dim I

o7 = 0

i8 : degree I  -- How many are there?

o8 = 4 

-- We compute them by elimination

i9 : eliminate (I, {l1,l2,y,z}) -- possible x values satisfy:

              4      2
o9 = ideal(36x  - 24x  + 1)  -- This means that y,z values are unique given x

i10 : eliminate (I, {l1,l2,z})  -- from this we get possible y values

               2        2
o10 = ideal (3y  - 1, 6x  + 6x*y - 1)

i11 : eliminate (I, {l1,l2}) -- and so on.

                                 2        2
o11 = ideal (x + y + z, 6y*z + 6z  - 1, 3y  - 1)

Answer 3

As Thomas commented, the function to consider is $$F=x^2+2 y^2+3 z^2+\lambda \left(x^2+y^2+z^2-1\right)+\mu (x+y+z)$$ Computing the derivatives $$F'_x=\mu +2 \lambda x+2 x$$ $$F'_y=\mu +2 \lambda y+4 y$$ $$F'_z=\mu +2 \lambda z+6 z$$ $$F'_\lambda=x^2+y^2+z^2-1$$ $$F'_\mu=x+y+z$$ From the first and second derivatives, we can eliminate $\lambda$ and $\mu$ to get (assuming $x\neq y$) $$\lambda=-\frac{x-2 y}{x-y}$$ $$\mu=-\frac{2 x y}{x-y}$$ Replacing in the third $$F'_z=-\frac{2 (x (y-2 z)+y z)}{x-y}$$ from which (assuming $2x\neq y$) $$z=\frac{x y}{2 x-y}$$ Replacing again $$F'_\lambda=x^2 \left(\frac{y^2}{(y-2 x)^2}+1\right)+y^2-1$$ $$F'_\mu=\frac{x y}{2 x-y}+x+y$$ So, we are left with $x,y$; the last equation allows to express $y=(1\pm \sqrt 3)x$. At this point, you are left with $F'_\lambda$ which will give you the solutions for $x$ for each possible $y$.

I am sure that you can easily take from here.

Answer 4

$x^2+2y^2+3z^2=x^2+y^2+z^2+y^2+2z^2=1+y^2+2z^2, x=-(y+z),(y+z)^2+y^2+z^2=1 \implies y^2+yz+z^2=\dfrac{1}{2}$, now the problem become : find min of $y^2+2z^2$ with $y^2+yz+z^2=\dfrac{1}{2}$,it shoud be easier now.

or we can go further :$y=\dfrac{-z \pm \sqrt{2-3z^2}}{2},f(z)=\left(\dfrac{-z \pm \sqrt{2-3z^2}}{2}\right)^2+2z^2$ which is single variable function.