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IF $x , y , z$ are arbitary positive real numbers satisfying the equation

$$ 4xy + 6yz + 8xz = 9$$

Find the maximum value of the product $xyz$.

I dont know from where to begin .

3 variables and one equation.

How I can achieve this?

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    $\begingroup$ Are you familiar with lagrange multipliers? I'm guessing not because of the tag, but just checking $\endgroup$ Mar 6, 2012 at 10:33
  • $\begingroup$ Lagrange Multipliers for constrained extremal points. $\endgroup$
    – agt
    Mar 6, 2012 at 10:34
  • $\begingroup$ @Alex Becker No.I am not aware of lagrange multipliers.If it is something related to my question.Please let me know $\endgroup$
    – vikiiii
    Mar 6, 2012 at 10:36
  • $\begingroup$ @vikiiii It's a technique used in calculus for exactly this kind of problem. $\endgroup$ Mar 6, 2012 at 10:38
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    $\begingroup$ This is the same sort of problem as the one in this other question of yours. The Wikipedia article on Lagrange multipliers shows you how to solve these; it has some worked out examples. $\endgroup$
    – joriki
    Mar 6, 2012 at 11:04

3 Answers 3

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Arithmetic mean is $\ge$ Geometric mean, i.e. $${{4xy + 6yz + 8xz}\over3} \ge {{(4xy\cdot 6yz\cdot 8xz)}}^{1/3}.$$

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  • $\begingroup$ how it will give me value of xyz? $\endgroup$
    – vikiiii
    Mar 6, 2012 at 11:01
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    $\begingroup$ LHS is AM of 18 things, RHS is not GM of 18 things. I think you get $(4xy+6yz+8xz)/3\ge((4xy)(6yz)(8xz))^{1/3}$. $\endgroup$ Mar 6, 2012 at 11:20
  • $\begingroup$ @GerryMyerson, you are right, i have edited the answer. $\endgroup$
    – quartz
    Mar 6, 2012 at 11:31
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    $\begingroup$ @vikiii RHS is $192^{1/3}*xyz^{2/3}$ and LHS is 3. $\endgroup$
    – quartz
    Mar 6, 2012 at 11:38
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    $\begingroup$ Of course, you then have to decide whether the bound you get is actually attained. $\endgroup$ Mar 6, 2012 at 11:49
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If we want to use the AGM inequality without loss we have to "symmetrize" the three variables. Therefore we replace $x$, $y$, $z$ by $$x':=4x,\quad y':=3y,\quad z':=6z\ .$$ The AGM inequality then says that $$\root 3\of {x'^2y'^2z'^2}\leq {x'y' + y'z' + z' x'\over 3}=4xy + 6 yz+8 zx=9\ ,$$ i.e., $x'y'z'\leq 27$, with equality iff $x'=y'=z'=3$. It follows that $$xyz = {x'y'z'\over 72}\leq{3\over 8}$$ with equality iff $x={3\over4}$, $y=1$, $z={1\over2}$.

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  • $\begingroup$ Thanks . But i want to know How did you think of x′:=4x,y′:=3y,z′:=6z . $\endgroup$
    – vikiiii
    Mar 7, 2012 at 4:40
  • $\begingroup$ @vikiiii: I wrote $x'=\alpha x$, $\ldots$, and determined $\alpha$, $\beta$, $\gamma$ such that $4xy={1\over3}x'y'$, $\ldots\ $. $\endgroup$ Mar 7, 2012 at 8:59
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You can also do it by using weighted arithmetic mean and geometric mean inequality,

$$\frac{m_1x_1 + m_2 x_2 + m_3 x_3 +...+ m_n x_n}{m_1+m_2+m_3....m_n} > [{x_1^{m_1}x_2^{m_2}x_3^{m_3}x_4^{m_4}....x_n^{m_n}}]^{\frac{1}{m_1m_1m_2m_3.....m_n}}$$

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