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Given $x,y,z >0$ and $xy^2z^3 = 108 $, what is the minimum value of $x+y+z$ ?

This is a homework problem, so if someone could just give me an outline/hint of the method used to solve this it would be much appreciated.

Thanks

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To prove the solution is $x=1,y=2,z=3$ we can write the constrained problem as $min f(y,z)$ where $f(y,z) = \frac{108}{y^2 z^3} + y + z$.

Solving for the minimum of $f(y,z)$ we get:

$$ \frac{\partial f}{\partial y} = -2 \frac{108}{y^3 z^3} +1 = 0 $$

$$ (yz)^3 = 216 $$

$$ yz = 6 $$

$$ \frac{\partial f}{\partial z} = -3 \frac{108}{y^2 z^4} +1 = 0 $$

$$ (yz^2)^2 = 324 $$

$$ yz^2 =18 $$

Solving the pair of quadratic equations we get $y=2$ and $z=3$, and hence $x=1$. So the minimum value of the sum is $6$.

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  • $\begingroup$ It remains to show that the point you've found is a minimum $\endgroup$
    – mrnovice
    May 2 '17 at 3:07
  • $\begingroup$ You are correct. Since the constraint space is compact, then any local max/min is a global max/min. And since the solution $x=108,y=1,z=1$ works and is bigger than $6$ we can conclude that $6$ is a minimum. $\endgroup$ May 2 '17 at 3:12
  • $\begingroup$ @ Emil Kerimov: Why is the constraint space compact ? The minimization of $f(y, z)$ requires $y > 0, z >0$ which is not a closed set. Secondly, the fact that f(1, 1) > f(2, 3), how does that help to conclude that it is a minimum ? $\endgroup$
    – me10240
    May 17 '19 at 3:18
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By AM-GM Inequality,$$x+y+z = x+\frac{y}{2}+\frac{y}{2}+\frac{z}{3}+\frac{z}{3}+\frac{z}{3}\geq6\left(\frac{xy^2z^3}{2^2\cdot3^3}\right)^\frac{1}{6}=6\left(\frac{108}{2^2\cdot3^3}\right)^\frac{1}{6}=6$$ Equality holds when $x=\frac{y}{2}=\frac{z}{3}$, so plugging this into $xy^2z^3=108$ we know the equality occurs when $x=1, y=2,z=3$.

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