## My Solution of The Purdue University Problem Of The Week No. 12

My solution of POW12 was accepted. [Remark of December 19, 2009: it is only stated that it was “completely or partially proved”]

Problem. Find, with proof, the maximum value of $\displaystyle\prod\limits_{j=1}^{k}x_{j}$ where $x_{j}\geq 0,\displaystyle\sum\limits_{j=1}^{k}x_{j}=100,$ and $k$ is variable. In particular, your answer must be greater than or equal to all values obtained from other choices of $k.$

Here is the solution I submited.

Solution.

Let $\left( x_{1},x_{2},\ldots ,x_{k}\right) \in\mathbb{R}^{k}$, $f\left( x_{1},x_{2},\ldots ,x_{k}\right) =\displaystyle\prod\limits_{j=1}^{k}x_{j}\in\mathbb{R}$, $c\left( x_{1},x_{2},\ldots ,x_{k}\right) =\displaystyle\sum_{j=1}^{k}x_{j}-100$ $\in\mathbb{R}$ and $\lambda\in\mathbb{R}$. For a given $k\in\mathbb{Z}$, with $k\geq 1$, we know by the Lagrange multipliers method that $f\left( x_{1}^{\ast },x_{2}^{\ast },\ldots ,x_{k}^{\ast }\right)$ is a local extremum if for $x^{\ast }=\left( x_{1}^{\ast },x_{2}^{\ast },\ldots,x_{k}^{\ast }\right)$

$\nabla f\left( x^{\ast }\right) -\nabla c\left( x^{\ast }\right) \lambda^{\ast }=0\quad \quad k\text{ equations}$

$c\left( x^{\ast }\right) =0\quad \quad 1\text{ equation}$

where $\lambda ^{\ast }$ is the value of the multiplier $\lambda$ that is a solution of these $k+1$ equations. Hence we have respectively

$\displaystyle\prod\limits_{j\neq i}^{k}x_{j}^{\ast }-\lambda ^{\ast }=0\quad\quad 1\leq i\leq k$

and

$\displaystyle\sum\limits_{j=1}^{k}x_{j}^{\ast}-100=0\text{. }$

Solving this system of equations we find

$x_{1}^{\ast }=x_{2}^{\ast }=\cdots =x_{i}^{\ast }\cdots =x_{k}^{\ast}=\lambda ^{\ast }=\dfrac{100}{k}$

and

$\displaystyle\prod\limits_{j=1}^{k}x_{j}^{\ast}=\left( \dfrac{100}{k}\right) ^{k}$,

the latter being a local extremum of $f\left( x_{1},x_{2},\ldots ,x_{k}\right)$, for a fixed $k$. We transformed the initial maximizing problem in $k$ continuous variables and the discrete variable $k$ into the maximizing of $\left( 100/k\right) ^{k}$. Now we introduce the following function

$u\left( t\right) =\left( \dfrac{100}{t}\right) ^{t}\quad\text{with }t\in\mathbb{R}\text{.}$

The function $u\left( t\right)$ has a maximum at the same point $t$ than the function

$v\left( t\right) =\ln u\left( t\right) =t\ln 100-t\ln t\text{.}$

On the other hand $v^{\prime }\left( t\right) =0$ for $t^{\ast }=e^{\ln100-1}\simeq 36.788,v^{\prime }\left( t\right) >0$ for $t and $v^{\prime }\left( t\right) <0$ for $t>t^{\ast }$. Therefore

$u\left( t^{\ast }\right) =\left( \dfrac{100}{t^{\ast }}\right) ^{t^{\ast }}$

is a maximum of $u\left( t\right)$. Since $u\left( 37\right) >u\left( 36\right)$, the maximum occurs at $k=37$. Thus, for $\sum_{j=1}^{k}x_{j}=\sum_{j=1}^{37}x_{j}=100$ we have

$\max\displaystyle\prod\limits_{j=1}^{k}x_{j}=\displaystyle\prod\limits_{j=1}^{37}x_{j}=\left( \dfrac{100}{37}\right) ^{37}\text{.}$

Update (Dec., 19,2009): some errors (identified by Rod Carvalho here) corrected.