The maximum value of the function $f (x, y, z) = xyz$

The maximum value of the function $$f (x, y, z) = xyz$$subject to the constraint $$xy+yz+zx-a =0, a >0$$ is

1. $a^{3/2}$
2. $\left(\dfrac{a}{3} \right) ^{3/2}$
3. $\left(\dfrac{3}{a} \right) ^{3/2}$
4. $\left(\dfrac{3a}{2} \right) ^{3/2}$

I am stuck on this problem. Can anyone help me please?

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What do you know to solve this problem? Do you know classical inequalities? Calculus? Lagrangian? What are your thoughts on this problem? –  Calvin Lin Dec 31 '12 at 7:23
@ Calvin Lin sir, I don't know where to begin......... –  Prasanta Dec 31 '12 at 7:25
@Prasanta Are $x,y,z$ given to be positive? –  user17762 Dec 31 '12 at 7:25
As you don't know where to start, I am giving you hint. Like in any other Optimization problems of such type, start with derivatives –  Ram Dec 31 '12 at 7:27
@Prasanta If $(x,y,z)$ are allowed to be negative, then the maximum is $\infty$, since $(x,y,z) = \left(-n,-n, \dfrac{n}2 - \dfrac{a}{2n} \right)$, which satisfies $xy+yz+zx = a$ and we get $xyz = \dfrac{n^3}2 - \dfrac{an}2$. Now let $n \to \infty$, to get the maximum as $\infty$. –  user17762 Dec 31 '12 at 7:32

Hint: I'll assume that $x,y,z>0$, since otherwise as Marvis points out, the maximum would be infinite. Applying the AM-GM inequality. We find that $$\left(\frac{xy+yz+zx}{3}\right)^3\geq (xyz)^2=f(x,y,z)^2.$$

Also, what happens when we set $$x=y=z=\left(\frac{a}{3}\right)^{\frac{1}{2}}.$$ Use this to complete the problem.

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Provided $x,y,z > 0$ –  user17762 Dec 31 '12 at 7:37
Your first statement is incorrect. See my comment to the question, where it is shown that if $(x,y,z)$ are allowed to be negative, then the maximum is in fact $\infty$. –  user17762 Dec 31 '12 at 7:41
@Marvis: Ahh, yes, thank you. I corrected my posted answer. –  Eric Naslund Dec 31 '12 at 7:42

By arithmetic-geometic inequality, $$\frac{xy+xz+yz}3\ge \sqrt[3]{xy\cdot xz\cdot yz}$$ provided $xy,xz,yz$ are all positive (i.e. $x,y,z$ have the same sign, e.g. are all positive) and with equality iff $xy=yz=xz$ (i.e. iff $x=y=z$). The left hand side is $\frac a3$ and the right hand side is $f(x,y,z)^{\frac23}$.

However, if $x,y,zh$ are allowed to have differnt signs, there is no absolute maximum: If $t>0$ and $y=z=-t$, then $x=t-\frac a{2t}$ produces $f(x,y,z)=t^3-\frac12at$, which becomes arbitrarily large.

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