3 variable multiplication with 1 constraint lagrange multiplier Using Lagrange multipliers, I need to calculate all points $(x,y,z)$ such that $$x^4y^6z^2$$ has a maximum or a minimum subject to the constraint that $$x^2 + y^2 + z^2 = 1$$
So, $f(x,y,z) = x^4y^6z^2 $
and $g(x,y,z) = x^2 + y^2 + z^2 - 1$
then i've done the partial derivatives
$$\frac{\partial f}{\partial x}(x,y,z)=\lambda\frac{\partial g}{\partial x}$$
which gives
$$4x^3y^6z^2 = 2xλ$$
$$6x^4y^5z^2 = 2yλ$$
$$2x^4y^6z = 2zλ$$
which i subsequently go on to find that 
$3x^2 = 2y^2 = 6z^2 $
This is where i've hit a dead end. Where do i go from here? or am i doing it all wrong? 
Thanks.
 A: The minimal value is $0$ for $x=0$.
The maximal value we can find by AM-GM:
$$x^4y^6z^2=108\left(\frac{x^2}{2}\right)^2\left(\frac{y^2}{3}\right)^3z^2\leq108\left(\frac{2\cdot\frac{x^2}{2}+3\cdot\frac{y^2}{3}+z^2}{6}\right)^6=\frac{1}{432}.$$
The equality occurs for $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$,
which says that $\frac{1}{432}$ is a maximal value.
Done!
If you wish to use the Lagrange multipliers method you need to add the following words.
Let $F(x,y,z,\lambda)=x^4y^6z^2+\lambda(x^2+y^2+z^2-1)$ and $A=\left\{(x,y,z,\lambda)|x^2+y^2+z^2=1\right\}$.
Sinse $F$ is a continuous function and $A$ is a compact, we see that $F$ gets on $A$ the maximal value and gets on $A$ the minimal value, which happens for solutions of your system.
The rest is to solve the system $\frac{x^2}{2}=\frac{y^2}{3}=z^2$ and $x^2+y^2+z^2=1$ for $xyz\neq0$ and to solve your system for $xyz=0$.
Now, you can choose, that you want. 
I think that to solve our problem by AM-GM is much better. 
A: You can now express $y^2$ and $z^2$ as functions of $x$ -- for example, $y^2=\frac32 x^2$.
Then, use the last equation you didn't use yet, which is
$$x^2+y^2+z^2=1$$
and plug in $y^2$ and $z^2$. You should get one single equation for $x$.
A: Lagrange multipliers, also called Lagrangian multipliers (e.g., Arfken 1985, p. 945), can be used to find the extrema of a multivariate function  subject to the constraint , where  and  are functions with continuous first partial derivatives on the open set containing the curve , and  at any point on the curve (where  is the gradient).
For an extremum of  to exist on , the gradient of  must line up with the gradient of . In the illustration above,  is shown in red,  in blue, and the intersection of  and  is indicated in light blue. The gradient is a horizontal vector (i.e., it has no -component) that shows the direction that the function increases; for  it is perpendicular to the curve, which is a straight line in this case. If the two gradients are in the same direction, then one is a multiple () of the other, so
(1)
The two vectors are equal, so all of their components are as well, giving
(2)
for all , ..., , where the constant  is called the Lagrange multiplier.
The extremum is then found by solving the  equations in  unknowns, which is done without inverting , which is why Lagrange multipliers can be so useful.
For multiple constraints , , ...,
To minimize a simple non-linear function by using Lagrange Multipliers.
EQUATION GIVEN :
5 - (x-2)^(2 )  - 2⋅(y-1)^2
x + 4y = 3
Consider the as f(x,y) and λ(x,y) 
f(x,y)  =  5 - (x-2)^(2 )  - 2⋅(y-1)^2
λ(x,y)  = x + 4y – 3
Adding equation 
g(x,y) = f(x,y) + λ(x,y)
∂g/∂x  = 0
∂g/∂y  = 0
∂g/∂λ  = 0
Taking derivative with respect to x
∂g/∂x  =  (∂(f(x,y) + λ(x,y)))/∂x
∂g/∂x  =  (∂(5 - (x-2)^(2 )  - 2⋅(y-1)^2  + x + 4y - 3))/∂x
∂g/∂x =    -2x + 4 + λ ……. [1]
Taking derivative with respect to y
∂g/∂y  =  (∂(f(x,y) + λ(x,y)))/∂y
∂g/∂y  =  (∂(5 - (x-2)^(2 )  - 2⋅(y-1)^2  + x + 4y - 3))/∂y
∂g/∂y =   - 4y + 4 + 4λ 
λ = y-1 ..........[2] 
Taking derivative with respect to λ
∂g/(∂λ )  =  (∂(f(x,y) + λ(x,y)))/(∂λ )
∂g/(∂λ )  =  (∂(5 - (x-2)^(2 )  - 2⋅(y-1)^2  + x + 4y - 3))/(∂λ )
∂g/(∂λ ) =   x + 4y - 3 ……[3]
Multiply [2] by 3, we get
9y - 3 = 0
y =  1/(3 )………[4]
Put the value of y in eq. [1], we get
λ = (1/(3 ) - 1)
λ  = (-2)/(3 )………. [5]
Put the value of y in eq. [3], we get
x + 4(1/3) - 3 = 0
x = -4(1/3) + 3 
x =  5/3………. [6]
f(x,y)  =  5 - (x-2)^(2 )  - 2⋅(y-1)^2
f(x,y)  =  5 - (5/3-2)^(2 )  - 2⋅(1/(3 )-1)^2
f(x,y) = 4
    Value of Constraint minimization is 4
