Lagrange multipliers: f(x,y,z)=xyz; x^2+2y^2+3z^2=6

[mathstresser]

Use Lagraange multipliers to find the maximum and minimum values of the function subject to the given constraint(s).

f(x,y,z)=xyz; x^2+2y^2+3z^2=6

So I did the partial derivatives for each of them and got

<yz,xz,xy>(lambda)

so

yz=xz, yz=xy, xz=xy

so

y=x, z=x, z=y

so

x=y=z

So, I substituted it into the equation and get

x^2+2x^2+3x^2=6 which equals

6x^2=6 x^2=1 x=+/-(1)

but, that is not the answer...
9
The answer is max= 2/(3)^(1/2)), min= -2/((3)^(1/2))

So, what am I doing wrong and what do I need to do to get the right answer?

 

[JakeD]

Use Lagraange multipliers to find the maximum and minimum values of the function subject to the given constraint(s).

f(x,y,z)=xyz; x^2+2y^2+3z^2=6

So I did the partial derivatives for each of them and got

<yz,xz,xy>(lambda)

so

yz=xz, yz=xy, xz=xy

so

y=x, z=x, z=y

so

x=y=z

So, I substituted it into the equation and get

x^2+2x^2+3x^2=6 which equals

6x^2=6 x^2=1 x=+/-(1)

but, that is not the answer...
9
The answer is max= 2/(3)^(1/2)), min= -2/((3)^(1/2))

So, what am I doing wrong and what do I need to do to get the right answer?

The Lagrangean is \(\displaystyle \L L(x,y,z,\lambda) = xyz - \lambda(x^2 + 2y^2 +3z^2 - 6).\) The partial derivatives of \(\displaystyle \L L\) are not what you say they are, e.g., \(\displaystyle \L \partial L/\partial y = xz - 4 \lambda y.\) It looks like you're on the right track, so try it again.