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I came across this question in a reference book:

$x,y,z$ are positive reals such that $x^3\cdot y^2\cdot z^4=17$.

Find the minimum value of $3x+5y+2z$.

Well, I know that the expression containing exponents will be greatest when $x/3=y/2=z/4$.

But how to minimize?

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    I think using Lagrange multipliers could help. – 2017-01-16
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    What's is that? – 2017-01-16

2 Answers 2

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Using weighted A. M G.M Inequality

$\displaystyle x+x+x+\frac{5y}{2}+\frac{5y}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}+\frac{z}{2}\geq 9\bigg(x\cdot x\cdot x\cdot \frac{5y}{2}\cdot \frac{5y}{2}\cdot \frac{z}{2}\cdot \frac{z}{2}\cdot \frac{z}{2}\cdot \frac{z}{2}\bigg)^{\frac{1}{9}}$

So $\displaystyle 3x+5y+2z\geq 9\bigg(x^3y^2z^4 \cdot \frac{25}{2^6}\bigg)^{\frac{1}{9}} = 9\left(\frac{17 \times 25}{2^6}\right)^{\frac{1}{9}}$

and equality hold when $\displaystyle x = \frac{5y}{2}=\frac{z}{2}$

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    What happend to 25 in the numerator? – 2017-01-16
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    This is not subject to the constraint that @Lokesh Sangewar provided. – 2017-01-16
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    @Ekesh yeah u are right – 2017-01-16
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    @Ekesh, Why not? The constraint is used in the second line! – 2017-01-16
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The method of lagrange multipliers is used in mathematical optimization when finding maxima and minima of a function subject to equality constraints.

There steps of using lagrange multipliers are:

  1. Solve the system of equations $\nabla f(x, y,z) = \lambda \nabla g(x, y,z)$, where $g$(x, y, z) is the constraint function and $f(x, y, z)$ is the function we are optimizing.
  2. Plug in the solutions and identify the maximum/minimums

In the question provided, we have $f(x, y, z) = 3x + 5y + 2z$ and $g(x) = x^3 \cdot y^2 \cdot z^4$.

We are also provided the constraint that $x, y, z > 0$.

We take the partial derivatives of $f(x)$ with respect to $x, y $ and $z$ and set them equal to the partial derivatives of $\lambda$ times the partial derivative of $g(x)$ with respect to $x, y$ and $z$ respectively to get the system of equations:

\begin{cases} 2 = 4x^3y^2z^3\lambda \\ 5 = 2x^3yz^4\lambda \\ 3 = 3x^2y^2z^4 \lambda \\ x^3 \cdot y^4 \cdot z^4 = 17 \end{cases}

Solving the system of equations yields that $\lambda$ can take many many values (Wolfram Alpha gave me around 10), but only one is real, which is 0.02599997.

This leads us to the conclusion that the minimum occurs when $x = \frac{z}{2}$ and when $x = \frac{5y}{2}$

We can substitute this into the constraint function and find that $x = 1.2341, y = 0.4936 $ and $z = 2.4682$.

Plugging these into $f(x, y, z)$ will give us the minimum value of approximately $f(x, y, z) = 11.1067$

Thank you to @Alex Silva for helping me with this.

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    Your solution is not correct because you do not take into account the constraints $x > 0, y >0$ and $z>0$. – 2017-01-16
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    "$x,y,z$ are positive reals ..." Read again! – 2017-01-16
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    But it is not the OP question! This not provide an answer for the question. – 2017-01-16
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    There were no negative solutions to the system of equations found from the lagrange multipliers; it would be impossible to have $x, y, z$ negative anyways. – 2017-01-16
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/51892/discussion-between-alex-silva-and-ekesh). – 2017-01-16