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

Posted on Dezembro 5, 2009 por

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

(Tradução aqui)

“ 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<t^{\ast } 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.

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Sobre Américo Tavares

eng. electrotécnico reformado / retired electrical engineer
Esta entrada foi publicada em Calculus, Cálculo, Matemática, Math, Problem, Problemas, Purdue University com as tags , . ligação permanente.

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