Lagrange Method Problem

[nikhilmaheshwari]

Maximize
f(x)=5x1+3x2f(x)=5x1+3x2
Subject to
g1(x)=x1+2x2+x3−6=0g1(x)=x1+2x2+x3−6=0
g2(x)=3x1+x2+x4−9=0g2(x)=3x1+x2+x4−9=0
x1,x2,x3≥0

 

[Deleted member 4993]

Maximize

f(x)=5x1+3x2f(x)=5x1+3x2​

Subject to
g1(x)=x1+2x2+x3−6=0g1(x)=x1+2x2+x3−6=0
g2(x)=3x1+x2+x4−9=0g2(x)=3x1+x2+x4−9=0
x1,x2,x3≥0

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[nikhilmaheshwari]

Hi, sorry but that question is given to us from our company person and I really don't know anything about Linear Problems, can you solve this for me..?

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520​

Please share your work/thoughts about this problem.

 

[Deleted member 4993]

Hi, sorry but that question is given to us from our company person and I really don't know anything about Linear Problems, can you solve this for me..?

Please ask that company person to contact us - with 'work' - so that we can know where to begin to help.....

 

[Steven G]

can you solve this for me..?

Nope, but we will help you solve it.

 

[JeffM]

By the way, it is a problem in linear programming and has nothing to do with differential equations.

 

[HallsofIvy]

You titled this "Lagrange Method Problem". Do you know what the "Lagrange Method" is?

The "Lagrange Method" (also called the "Lagrange Multipler Method") for minimizing (or maximizing) the function $\displaystyle F(x_1, x_2, x_3)$ subject to constraints $\displaystyle g_1(x_1, x_2, x_3)= A$, $\displaystyle g_2(x_1, x_2, x_3)= B$, etc. is to minimize (or maximize) $\displaystyle F(x_1, x_2)- \lambda_1 g_1(x_1, x_2, x_3)- \lambda_2 g_2(x_1, x_2, x_3)+ \cdot\cdot\cdot$. Take the derivatives with respect to $\displaystyle x_1$, $\displaystyle x_2$, $\displaystyle x_3$ and all of the "$\displaystyle \lambda$"s and set them equal to 0. With n constraints, so n "$\displaystyle \lambda$"s, that gives n+ 3 equations to solve for $\displaystyle x_1$,$\displaystyle x_2$, $\displaystyle x_3$, and the "$\displaystyle \lambda$"s. Since specific values for the "$\displaystyle \lambda$"s is not necessary for a solution, it is often simplest to eliminate them by dividing one equation by another.

 

[HallsofIvy]

You might ask your "company person" to clarify the question. Your function has only $\displaystyle x_1$ and $\displaystyle x_2$ while your constraints also involve $\displaystyle x_4$ but you have conditions that $\displaystyle x_1$, $\displaystyle x_2$, and $\displaystyle x_3$ are non-negative with no mention of $\displaystyle x_4$! I am going to assume that the "$\displaystyle x_4$" in the second constraint was supposed to be "$\displaystyle x_3$.

Here the function is $\displaystyle F(x_1, x_2, x_3)= 5x_1+ 3x_2$ (no "$\displaystyle x_3$"?) and the constraints are $\displaystyle g_1(x_1, x_2, x_3)= x_1+ 2x_2+ x_3= 6$ and
$\displaystyle g_2(x_1, x_2, x_3)= 3x_1+ x_2+ x_3= 9$.

So $\displaystyle F(x_1, x_2, x_3)= 5x_1+ 3x_2+ \lambda_1(x_1+ x_2+ x_3)+ \lambda_2(3x_1+ x_2+ x_3)$.

The derivative with respect to $\displaystyle x_1$ is $\displaystyle 5+ \lambda_1+ 3\lambda_2= 0$.
The derivative with respect to $\displaystyle x_2$ is $\displaystyle 3+ \lambda_1+ 2\lambda_2= 0$.
The derivative with respect to $\displaystyle x_3$ is $\displaystyle \lambda_1+ \lambda_2= 0$.
The derivative with respect to $\displaystyle \lambda_1$ is $\displaystyle x_1+ x_2+ x_3= 0$.
The derivative with respect to $\displaystyle \lambda_2 \ is \ 3x_1+ x_2+ x_3= 0$.

Solve those four equations for $\displaystyle x_1$, $\displaystyle x_2$. You could, of course, solve for $\displaystyle \lambda_1$ and $\displaystyle \lambda_2$ but those are not asked.[/tex]