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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?2012-12-31
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    @ Calvin Lin sir, I don't know where to begin.........2012-12-31
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    @Prasanta Are $x,y,z$ given to be positive?2012-12-31
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    As you don't know where to start, I am giving you hint. Like in any other Optimization problems of such type, start with derivatives2012-12-31
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    @ Marvis sir, sign of $x,y,z$ is not given in the question.......2012-12-31
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    @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 $\infty$.2012-12-31
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    I don't get why these questions by Prasanta are getting down voted. You may think that they are silly/easy, but OP doesn't know how to approach them AND is willing to work through these with some guidance. For that, he deserves to be up voted. Take a look at http://math.stackexchange.com/questions/268029/trace-of-the-matrix-i-m-m2-is for example.2012-12-31
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    @Marvis: Dear Marvis I know that if we consider $F(x,y,z)=0$ subject to constraint condition $G(x,y,z)=0$ then the necessary condition that $F$ have an extreme is $F_xG_y-F_yG_x=0$. As I did this didn't hold for this problem. It seems strange to me or I am missing something? Thanks.2012-12-31
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    @BabakSorouh How do you get the condition $F_x G_y - F_y G_x = 0$?2012-12-31
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    @Marvis: It is a solved problem in Advanced Calculus by M.R.Spiegel. He proved that there.2012-12-31

2 Answers 2

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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$2012-12-31
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    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 $\infty$.2012-12-31
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    @Marvis: Ahh, yes, thank you. I corrected my posted answer.2012-12-31
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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.