I'm trying to find the max and mins of the equation $f(x,y,z) = xy + 3xz + 2yz$ on the constraint, $g(x,y,z)=5x+9y+z-10$. So according to the Lagrange Multiplier procedure, I take the partial derivatives of both equations and get,
$\frac{\partial f}{\partial x} = y + 3z$, $\frac{\partial g}{\partial x} = 5$
$\frac{\partial f}{\partial y} = x + 2z$, $\frac{\partial g}{\partial y} = 9$
$\frac{\partial f}{\partial z} = 3x + 2y$, $\frac{\partial g}{\partial z} = 1$
Then, using $\nabla f = \lambda\nabla g$
$y+3z = 5\lambda$
$x+2z = 9\lambda$
$3x+2y = \lambda$
However, the problem I encounter now is that I am having trouble getting the variables x, y, and z in terms of one variable when solving for lambda. If I am unable to do so, does this mean that the maxima and minima are undefined? I also noticed that the constraint was a plane, so perhaps there can't be any extrema because the constraint equation runs on values of x, y, and z that can take range from $\infty$ to $-\infty$ (no restriction on domain)?
One approach is to multiply the first equation by 9, the second by 5, the third by 45, giving all 3 equations a RHS of $45\lambda$, so we can set them equal to each other.
$9y+27z=45\lambda$
$5x+10z=45\lambda$
$135x+90y=45\lambda$
Set the first equal to the second, the second equal to the third, to give:
$9y+27z=5x+10z$
$5x+10z=135x+90y$
Now, simplify to the following:
$17z=5x-9y$
$10z=130x+90y$
Divide the second equation by 10 to get the pair of equations
$17z=5x-9y$
$z=13x+9y$
Now, try adding the equations, and substituting the result into one of the equations.
P.S. When approaching Lagrange Multipliers, I tend to avoid computing the value of $\lambda$ unless absolutely necessary, only resorting to this when literally everything else fails. One other thing: I have assumed $\lambda \neq 0$ throughout. Indeed, if $\lambda=0$, you can show that $x=y=z=0$ is the only solution to those 3 original equations, which doesn't satisfy the constraint.