# 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?

• i think there is no maximum or minimum. – Dr. Sonnhard Graubner Mar 28 '18 at 18:21
• Sorry I used a wrong phrase. It was meant to find the biggest value. – Eric Mar 28 '18 at 18:28
• are $x$, $y$, and $z$ constrained to be positive? – K B Dave Mar 28 '18 at 18:40

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 $\frac{343}{6}.$

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.

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.