MHB Is This Solution to the Linear Programming Problem Correct?

AI Thread Summary
The discussion revolves around solving a linear programming problem with the objective function to maximize \(5x_1 - 4x_2\) under specific constraints. One participant claims to have found the solution as \((0, \frac{6}{5}, \frac{36}{5}, 0, \frac{99}{5})\), while another participant suggests a different solution of \(x_1=12\) and \(x_2=6\) with a maximum value of 36. There is also a suggestion to solve the problem graphically due to the simplicity of having only two variables. The conversation highlights discrepancies in the proposed solutions and encourages exploring graphical methods for verification.
evinda
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Hello! (Wave)

I want to solve the linear programming problem:
$\max (5x_1-4x_2) \\ -x_1+x_2 \geq -6 \\ 3x_1-2x_2 \leq 24 \\ -2x_1+3x_2 \leq 9 \\ x_1, x_2 \geq 0$

I have found that the solution is $\left(0, \frac{6}{5}, \frac{36}{5},0, \frac{99}{5} \right)$.. Am I right?
 
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evinda said:
Hello! (Wave)

I want to solve the linear programming problem:
$\max (5x_1-4x_2) \\ -x_1+x_2 \geq -6 \\ 3x_1-2x_2 \leq 24 \\ -2x_1+3x_2 \leq 9 \\ x_1, x_2 \geq 0$

I have found that the solution is $\left(0, \frac{6}{5}, \frac{36}{5},0, \frac{99}{5} \right)$.. Am I right?

Hey evinda! (Smile)

I'm getting $x_1=12,\ x_2=6,\ \max=36$. (Worried)
 
evinda said:
Hello! (Wave)

I want to solve the linear programming problem:
$\max (5x_1-4x_2) \\ -x_1+x_2 \geq -6 \\ 3x_1-2x_2 \leq 24 \\ -2x_1+3x_2 \leq 9 \\ x_1, x_2 \geq 0$

I have found that the solution is $\left(0, \frac{6}{5}, \frac{36}{5},0, \frac{99}{5} \right)$.. Am I right?

As there are only two variables you could solve this graphically. Have you tried this?
 
I was reading a Bachelor thesis on Peano Arithmetic (PA). PA has the following axioms (not including the induction schema): $$\begin{align} & (A1) ~~~~ \forall x \neg (x + 1 = 0) \nonumber \\ & (A2) ~~~~ \forall xy (x + 1 =y + 1 \to x = y) \nonumber \\ & (A3) ~~~~ \forall x (x + 0 = x) \nonumber \\ & (A4) ~~~~ \forall xy (x + (y +1) = (x + y ) + 1) \nonumber \\ & (A5) ~~~~ \forall x (x \cdot 0 = 0) \nonumber \\ & (A6) ~~~~ \forall xy (x \cdot (y + 1) = (x \cdot y) + x) \nonumber...
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