Global maximum and minimum of $f(x,y,z)=24xyz$ with the constraint $8x+3y+6z=21$? I have to find  $f(x,y,z)=24xyz$ maximum value on surface $8x+3y+6z=21$. I have found the partial derivatives of $f$ and used Langranian to form 
$$8x+3y+6z=21\\
24yz=\lambda 8\\
24xz=\lambda 3\\
24xy=\lambda 6$$
So, my problem is that how can I solve the points $x$,$y$,$x$ and Langrange multiplier $\lambda$ to find the maximum point?
Thanks for your answers in advance!
 A: There is no global maximum since setting say $x=-1$, $y=\frac{29}{3}-2z$ (which implies that the constraint is satisfied), then
$$f(x,y,z)=24z(2z-\frac{29}{3}),$$
and the function can be made arbitrarily large by making $z$ arbitrarily large.
There is also no global minimum since setting say $x=1$, $y=\frac{13}{3}-2z$ (which implies that the constraint is satisfied), then
$$f(x,y,z)=24z(\frac{13}{3}-2z),$$
and the function can be made arbitrarily negative by making $z$ arbitrarily large.
A: It is natural to assume that $x,y,z$ are non-negative. Then we are restricted to a compact set, so the extrema do exist by the extreme value theorem. By the AM-GM inequality we have $$\frac{8x+3y+6z}{3}\ge\sqrt[3]{144xyz},$$ so $$144xyz\le 7^3.$$ This means that $$24xyz\le\frac{343}{6}.$$ The equality we have for $8x=3y=6z=7$. So, the maximal value is $\frac{343}{6}.$
A: Multiplying each equation by $x,y,z$ respectively gives:
$24xyz = 8\lambda x  = 3\lambda y = 6\lambda z$ 
And the constraint says:
$8x + 3y + 6z = 21\\
8x = 3y = 6z = 7\\
24xyz = 24\frac {7}{8}\frac {7}{3}\frac {7}{6} = \frac {343}{6}$
But this is just a local maximum.
If we consider a value of $x,$ and a moderate positive value of $y$, and a very negative value of $z,$ we satisfy the constraint and $f(x,y,z)$ is very negative.
And, if we consider a value of $x,$ and a moderate negative value of $y$, and a very negative value of $z,$  $f(x,y,z)$ is large.
It should be clear that $f(x,y,z)$ is unbounded despite the constraint.  
