MHB Possible title: Linear Optimization Problem: Finding Optimal Solutions

[goosey00]

Linear Optimization Problem follow up

Maximize: z=2x₂+5x₂+x₃
x₁+x₂+x₃ less then or equal to 12
x₁-x₂ less then or equal to 15
x₂+2x₃ less then or equal to 10
x1, x₂ and x₃ is greater then or equal to 0
x₁= x₂= x₃= s₁= s₂= s₃= z=
I get x₁=2, x₂=10 x₃=0 s₁=0 s₂=0 s₃=0 z=0
Is this wrong??

 

[Opalg]

Re: Linear Optimization Problem follow up

Maximize: z=2x₂+5x₂+x₃ I assume that the first x₂ should be x₁.
x₁+x₂+x₃ less then or equal to 12
x₁-x₂ less then or equal to 15
x₂+2x₃ less then or equal to 10
x1, x₂ and x₃ is greater then or equal to 0
x₁= x₂= x₃= s₁= s₂= s₃= z=
I get x₁=2, x₂=10 x₃=0 s₁=0 s₂=0 s₃=0 z=0
Is this wrong??

It looks as though $x_1=2,\ x_2=10,\ x_3=0$ is correct, but why do you get $z=0$? The condition $z=2x_1+5x_2+x_3$ says that $z$ should be 54 for those values of the $x$'s.

 

[goosey00]

Re: Linear Optimization Problem follow up

It looks as though $x_1=2,\ x_2=10,\ x_3=0$ is correct, but why do you get $z=0$? The condition $z=2x_1+5x_2+x_3$ says that $z$ should be 54 for those values of the $x$'s.

So where do you get 54?
BTW-you are right, it was suppose to be 1.

 

[Jameson]

Re: Linear Optimization Problem follow up

So where do you get 54?
BTW-you are right, it was suppose to be 1.

If you plug in $x_1=2$, $x_2=10$ and $x_3=0$ into $z=2x_1+5x_2+x_3$ you get $z=(2 \times 2) + (5 \times 10) + 0=54$.

 

[Sudharaka]

Re: Linear Optimization Problem follow up

Maximize: z=2x₂+5x₂+x₃
x₁+x₂+x₃ less then or equal to 12
x₁-x₂ less then or equal to 15
x₂+2x₃ less then or equal to 10
x1, x₂ and x₃ is greater then or equal to 0
x₁= x₂= x₃= s₁= s₂= s₃= z=
I get x₁=2, x₂=10 x₃=0 s₁=0 s₂=0 s₃=0 z=0
Is this wrong??

Hi goosey00, :)

You can check the optimal solutions of linear programming problems >>here<<.

Kind Regards,
Sudharaka.