# Math Help - More Details on The Optimization Problem

1. ## More Details on The Optimization Problem

Hello,

I need the solution of the following problem:

max/min $f(x_{1},x_{2},...,x_{n})= \frac{x_{1}}{1-x_{n}}$ subject to $x_{1}+x_{2}+...+x_{n}=1.$

where $0

Best Regards.

2. Originally Posted by raed
Hello,

I need the solution of the following problem:

max/min $f(x_{1},x_{2},...,x_{n})= \frac{x_{1}}{1-x_{n}}$ subject to $x_{1}+x_{2}+...+x_{n}=1.$

where $0

Best Regards.
If there are no mistakes $f= \frac{x_{1}}{1-x_{n}}$ seems to be function only of $x_{1}$ and $x_{n}$ and not of $x_{2}$, $x_{3}$,..., $x_{n-1}$...

Kind regards

$\chi$ $\sigma$

3. Originally Posted by chisigma
If there are no mistakes $f= \frac{x_{1}}{1-x_{n}}$ seems to be function only of $x_{1}$ and $x_{n}$ and not of $x_{2}$, $x_{3}$,..., $x_{n-1}$...

Kind regards

$\chi$ $\sigma$
Thank you very much for your reply, there is no mistake because in my own problem I will maximize/minimize functions such the one above but in each case $x_{1}$ will be replaced by $x_{2}$, $x_{3}$,..., $x_{n-1}$

Best Regards,

4. The function $f$ belongs to $\mathcal{C}^1(A)$ where $A$ is the open set $A=\prod_{i=1}^n(a_i,b_i)$ .

On the other hand, $g(x_1,\ldots,x_n)=x_1+\ldots+x_n-1$ also belongs to $\mathcal{C}^1(A)$ and $\textrm{rank}(\;\nabla g(x_1,\ldots,x_n)\=1" alt="\textrm{rank}(\;\nabla g(x_1,\ldots,x_n)\=1" /> for all $x=(x_1,\ldots,x_n)\in A$ (maximun rank).

According to the multipliers Lagrange theorem if $f$ has a local maximum or minimum at $x_0\in A$ there exists $\lambda\in \mathbb{R}$ such that $x_0$ is critical point of

$F(x_1,\ldots,x_n)=f(x_1,\ldots,x_n)+\lambda g(x_1,\ldots,x_n)$

Verify that such a point does not exist. This means that the absolute maximum an minimum value for $f$ is in the boundary $\partial (K)$ of the compact set:

$K=\left(\;\prod_{i=1}^n[a_i,b_i]\;\right)\cap\{(x_1,\ldots,x_n):x_1+\ldots+x_n=1\}$

because $f:K\to \mathbb{R}$ is continuous.

5. We can write

$
\displaystyle
f=\frac{x_1}{1-x_n}=\frac{x_1}{\sum_1^{n-1}x_i}
$

If $x_i$ are independent in their range

$
\displaystyle
max \; f=\frac{max(x_1)}{min(\sum_1^{n-1}x_i)}=\frac{b_1}{\sum_1^{n-1}a_i}
$

and in the similar way

$
\displaystyle
min \; f=\frac{min(x_1)}{max(\sum_1^{n-1}x_i)}=\frac{a_1}{\sum_1^{n-1}b_i}
$

6. Originally Posted by FernandoRevilla
The function $f$ belongs to $\mathcal{C}^1(A)$ where $A$ is the open set $A=\prod_{i=1}^n(a_i,b_i)$ .

On the other hand, $g(x_1,\ldots,x_n)=x_1+\ldots+x_n-1$ also belongs to $\mathcal{C}^1(A)$ and $\textrm{rank}(\;\nabla g(x_1,\ldots,x_n)\=1" alt="\textrm{rank}(\;\nabla g(x_1,\ldots,x_n)\=1" /> for all $x=(x_1,\ldots,x_n)\in A$ (maximun rank).

According to the multipliers Lagrange theorem if $f$ has a local maximum or minimum at $x_0\in A$ there exists $\lambda\in \mathbb{R}$ such that $x_0$ is critical point of

$F(x_1,\ldots,x_n)=f(x_1,\ldots,x_n)+\lambda g(x_1,\ldots,x_n)$

Verify that such a point does not exist. This means that the absolute maximum an minimum value for $f$ is in the boundary $\partial (K)$ of the compact set:

$K=\left(\;\prod_{i=1}^n[a_i,b_i]\;\right)\cap\{(x_1,\ldots,x_n):x_1+\ldots+x_n=1\}$

because $f:K\to \mathbb{R}$ is continuous.
Thank you very much for your reply, but the function $f$ belongs to $\mathcal{C}^1(A)$ where $A$ is the closed set $A=\prod_{i=1}^n[a_i,b_i]$ not the open set $A=\prod_{i=1}^n(a_i,b_i)$ .

7. Originally Posted by zzzoak
We can write

$
\displaystyle
f=\frac{x_1}{1-x_n}=\frac{x_1}{\sum_1^{n-1}x_i}
$

If $x_i$ are independent in their range

$
\displaystyle
max \; f=\frac{max(x_1)}{min(\sum_1^{n-1}x_i)}=\frac{b_1}{\sum_1^{n-1}a_i}
$

and in the similar way

$
\displaystyle
min \; f=\frac{min(x_1)}{max(\sum_1^{n-1}x_i)}=\frac{a_1}{\sum_1^{n-1}b_i}
$
Thank you very much for reply, accourding to my problem the $x_i$ are dependent in their range.

Best Regards,

8. Originally Posted by raed
Hello,

I need the solution of the following problem:

max/min $f(x_{1},x_{2},...,x_{n})= \frac{x_{1}}{1-x_{n}}$ subject to $x_{1}+x_{2}+...+x_{n}=1.$

where $0

Best Regards.
There are only two variables $x_1$ and $x_n$, and the function has no calculus like minimum or maximum in the interior of the feasible region.

As the feasible region is closed there is a minimum/maximum on the boundary and it can be found by examination of the function on the boundary.

CB

9. Originally Posted by raed
Thank you very much for your reply, but the function $f$ belongs to $\mathcal{C}^1(A)$ where $A$ is the closed set $A=\prod_{i=1}^n[a_i,b_i]$ not the open set $A=\prod_{i=1}^n(a_i,b_i)$ .

Look at my previous post, I took the restriction to $A=\textrm{Int}(K)$ to prove that we get the absolute maximum and minimum on the boundary of $K$.

10. Originally Posted by FernandoRevilla
Look at my previous post, I took the restriction to $A=\textrm{Int}(K)$ to prove that we get the absolute maximum and minimum on the boundary of $K$.
Dear Prof. Fernando Revilla,

I understand you but what is the solution of the problem in this case.

Best Regards,

11. Originally Posted by raed
I understand you but what is the solution of the problem in this case.

First of all, you should show some work. Look rule #11 here:

http://www.mathhelpforum.com/math-he...ng-151424.html

I propose you the following: consider $n=3$ (then, you'll analyze what difficulties you can find in the general case).

(i) Prove that there are no critical points in $\textrm{Int}(K)$ for:

$F(x,y,z)=\dfrac{x}{1-z}+\lambda (x+y+z-1)$

(ii) Find the absolute maximum and minimum of $f:\partial (K)\to \mathbb{R}$

$f(x,y,z)=\dfrac{x}{x+y},\quad \partial (K) \equiv \begin{Bmatrix} a_1\leq x \leq b_1\\ a_2 \leq y \leq b_1\\a_3\leq 1-x-y \leq b_3 \end{matrix}\quad (\; [a_i,b_i]\subset (0,1)\ " alt="f(x,y,z)=\dfrac{x}{x+y},\quad \partial (K) \equiv \begin{Bmatrix} a_1\leq x \leq b_1\\ a_2 \leq y \leq b_1\\a_3\leq 1-x-y \leq b_3 \end{matrix}\quad (\; [a_i,b_i]\subset (0,1)\ " />

Draw in the $xy$ plane the set $\partial(K)$ . According to the relations of $a_3,b_3$ with $a_1,b_1,a_2,b_2$ , analyze the different kind of boundaries you can find.

P.S. Have you seen the problem in a book? . In what a context?. Is it homework?.

12. Dear Prof. Fernando Revilla,

First of all thank you very much for every thing.
My problem is neither in a book nor a homework. Nowadays I am working on a paper and I countered this optimization problem. I think if I was able to sove it, then I would get some results concerning my research. Therefore, I first introduced the problem in a general case without determining the function and the constraint.

Best Regards.

13. Originally Posted by raed
My problem is neither in a book nor a homework. Nowadays I am working on a paper and I countered this optimization problem. I think if I was able to sove it, then I would get some results concerning my research. Therefore, I first introduced the problem in a general case without determining the function and the constraint.

Well, in that case and if we have no more information about the parameters $a_i,b_i$ this could be a large, large,...., large problem with a lot of cases to analyze depending on the form of the boundary.