Hi guys I am working with the following polynomial and I am trying to find the $\lambda , \mu$. I have a polynomial and I am trying to do Lagrange multipliers. Here is what I have.
$f(x,y,z)= a \log(x)+ b \log(y)+c \log(z)$ with the constraints $x+y+z=1$ and $xz-y^2=0$. ($a,b,c$ are just constants) So I made $F(x,y,z)=a \log(x)+ b \log(y)+c \log(z)+ \lambda(x+y+z-1) + \mu(xz-y^2)$ took the partial derivatives with respect to x,y,z,$\lambda, \mu$ and go
$$F_x= \frac{a}{x}+\lambda +\mu (z)$$
$$F_y= \frac{b}{y}+\lambda +\mu (-2y)$$
$$F_z= \frac{c}{z}+\lambda +\mu (x)$$
Then I set them all equal to zero and clear the denominators. Then added the equations to get: $$0 =a+b+c+ 3 \lambda + 2 \mu (xz-y^2)$$ using the second constraint I get that $$\lambda =-(a+b+c)/3$$
That is great it is sort of what I expected but $\mu$ is problematic to find it I thought now I plug back into the $F_x,F_y,F_z$ the $\lambda$ now I am unsure what to do I was thinking if I add the equations without the clearing the denominator I might something but it does not seem simple. I would appreciate someones input, or another set of eyes
Also If someone knows how to do this in macaulay 2 I would really like to learn how to do this
You can get rid of one constraint for free: All three variables $x$, $y$, $z$ have to be positive. It is therefore allowed to put $$x:=u^2,\quad y:=uv,\quad z:=v^2\ ,$$ so that $xz=y^2$ is automatically satisfied. We now have to investigate the function $$g(u,v):=(2a+b)\log u+(2c+b)\log v$$ under the sole constraint $u^2+uv+v^2=1$. We obtain the conditions $$G_u={2a+b\over u}-\lambda(2u+v)=0,\quad G_v={2c+b\over v}-\lambda(u+2v)=0\ .$$ Multiplying the first equation by $u$, the second by $v$, and adding gives $$2(a+b+c)-2\lambda(u^2+uv+v^2)=0\ ,$$ whence $\lambda=a+b+c$. Etcetera.