Prove that $(2,0,0)$ is the optimal solution to this problem.

P) Minimize $2x_1+5x_2+7x_3$ subject to constraints:
$7x_1+6x_2+3x_3-s_1=14$
$2x_1+4x_2+5x_3+s_2=4$
Where: $x_1,x_2,x_3 \ge 0$

This question asked me to prove me that $(2,0,0)$ is the optimal solution for the primal problem Without using simplex.(using complementary slackness conditions)

The dual is: Maximize $14y_1+4y_2$
$7y_1+2y_2-t_1=2$
$6y_1+4y_2-t_2=5$
$3y_1+5y_2-t_3=7$
Where: $y_1 \ge 0,y_2 \le 0$

Now if we use complementary slackness theorem: $x_1>0$ then $t_1=0 $
The equation will be: $7y_1+2y_2=2$

Then:

$x_2=0 , x_3=0$ then $t_2,t_3=?$
(We cannot determine $t_2, t_3$)

If we substitute $(2,0,0)$ in the primal we see $s_1,s_2=0$
And we cannot determine $y_1$ and $y_2$

So we only get one equation $7y_1+2y_2=2$.

Now should i say we can't prove that $(2,0,0)$ is the optimal answer for the problem? Or am i wrong?

Prove that $(2,0,0)$ is the optimal solution to this problem.

P) Minimize $2x_1+5x_2+7x_3$ subject to constraints:
$7x_1+6x_2+3x_3-s_1=14$
$2x_1+4x_2+5x_3+s_2=4$
Where: $x_1,x_2,x_3 \ge 0$

This question asked me to prove me that $(2,0,0)$ is the optimal solution for the primal problem Without using simplex.(using complementary slackness conditions)

The dual is: Maximize $14y_1+4y_2$
$7y_1+2y_2+t_1=2$
$6y_1+4y_2+t_2=5$
$3y_1+5y_2+t_3=7$
Where: $y_1 \ge 0,y_2 \le 0$

Now if we use complementary slackness theorem: $x_1>0$ then $t_1=0 $
The equation will be: $7y_1+2y_2=2$

Then:

$x_2=0 , x_3=0$ then $t_2,t_3=?$
(We cannot determine $t_2, t_3$)

If we substitute $(2,0,0)$ in the primal we see $s_1,s_2=0$
And we cannot determine $y_1$ and $y_2$

So we only get one equation $7y_1+2y_2=2$.

Now should i say we can't prove that $(2,0,0)$ is the optimal answer for the problem? Or am i wrong?

Prove that $(2,0,0)$ is the optimal solution to this problem.

P) Minimize $2x_1+5x_2+7x_3$ subject to constraints:
$7x_1+6x_2+3x_3-s_1=14$
$2x_1+4x_2+5x_3+s_2=4$
Where: $x_1,x_2,x_3 \ge 0$

This question asked me to prove me that $(2,0,0)$ is the optimal solution for the primal problem Without using simplex.(using complementary slackness conditions)

The dual is: Maximize $14y_1+4y_2$
$7y_1+2y_2-t_1=2$
$6y_1+4y_2-t_2=5$
$3y_1+5y_2-t_3=7$
Where: $y_1 \ge 0,y_2 \le 0$

Now if we use complementary slackness theorem: $x_1>0$ then $t_1=0 $
The equation will be: $7y_1+2y_2=2$

Then:

$x_2=0 , x_3=0$ then $t_2,t_3=?$
(We cannot determine $t_1 , t_2$)

If we substitute $(2,0,0)$ in the primal we see $s_1,s_2=0$
And we cannot determine $y_1$ and $y_2$

So we only get one equation $7y_1+2y_2=2$.

Now should i say we can't prove that $(2,0,0)$ is the optimal answer for the problem? Or am i wrong?

Prove that $(2,0,0)$ is the optimal solution to this problem.

P) Minimize $2x_1+5x_2+7x_3$ subject to constraints:
$7x_1+6x_2+3x_3-s_1=14$
$2x_1+4x_2+5x_3+s_2=4$
Where: $x_1,x_2,x_3 \ge 0$

This question asked me to prove me that $(2,0,0)$ is the optimal solution for the primal problem Without using simplex.(using complementary slackness conditions)

The dual is: Maximize $14y_1+4y_2$
$7y_1+2y_2-t_1=2$
$6y_1+4y_2-t_2=5$
$3y_1+5y_2-t_3=7$
Where: $y_1 \ge 0,y_2 \le 0$

Now if we use complementary slackness theorem: $x_1>0$ then $t_1=0 $
The equation will be: $7y_1+2y_2=2$

Then:

$x_2=0 , x_3=0$ then $t_2,t_3=?$
(We cannot determine $t_2, t_3$)

If we substitute $(2,0,0)$ in the primal we see $s_1,s_2=0$
And we cannot determine $y_1$ and $y_2$

So we only get one equation $7y_1+2y_2=2$.

Now should i say we can't prove that $(2,0,0)$ is the optimal answer for the problem? Or am i wrong?