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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?

  • 0
    @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.

  • 0
    @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.