Mixing
Is funtion Is function $x_1+x_2+x_3+x_4$ bounded on$D:={(x_1,x_2,x_3,x_4) | x_1-x_2-3x_3+x_4 = 5,...
$begingroup$
Is function $x_1+x_2+x_3+x_4$ bounded on the domain $D:={(x_1,x_2,x_3,x_4) | x_1-x_2-3x_3+x_4 = 5, 2x_1+2x_2+x_3-2x_4 = 4,& quad forall i: x_i geq 0}$
I have no idea on how to find an answer. Can anyone give me some hints?
calculus linear-algebra optimization linear-programming
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
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$begingroup$
Is function $x_1+x_2+x_3+x_4$ bounded on the domain $D:={(x_1,x_2,x_3,x_4) | x_1-x_2-3x_3+x_4 = 5, 2x_1+2x_2+x_3-2x_4 = 4,& quad forall i: x_i geq 0}$
I have no idea on how to find an answer. Can anyone give me some hints?
calculus linear-algebra optimization linear-programming
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
$endgroup$
add a comment |
$begingroup$
Is function $x_1+x_2+x_3+x_4$ bounded on the domain $D:={(x_1,x_2,x_3,x_4) | x_1-x_2-3x_3+x_4 = 5, 2x_1+2x_2+x_3-2x_4 = 4,& quad forall i: x_i geq 0}$
I have no idea on how to find an answer. Can anyone give me some hints?
calculus linear-algebra optimization linear-programming
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
$endgroup$
Is function $x_1+x_2+x_3+x_4$ bounded on the domain $D:={(x_1,x_2,x_3,x_4) | x_1-x_2-3x_3+x_4 = 5, 2x_1+2x_2+x_3-2x_4 = 4,& quad forall i: x_i geq 0}$
I have no idea on how to find an answer. Can anyone give me some hints?
calculus linear-algebra optimization linear-programming
calculus linear-algebra optimization linear-programming
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
edited Feb 19 at 6:13
Thomas Edison
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
asked Jan 9 '17 at 13:17
Thomas EdisonThomas Edison
237314
237314
add a comment |
add a comment |
1 Answer
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No. Let $x_3 = 0$ then, $x_1-x_2 = 5-x_4, x_1+x_2 = x_4+2$. So, $x_1 = 3.5, x_2 = x_4-1.5$ and $x_1+x_2+x_3+x_4 = 2x_4+2 $ is unbounded.
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
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1 Answer
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1 Answer
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$begingroup$
No. Let $x_3 = 0$ then, $x_1-x_2 = 5-x_4, x_1+x_2 = x_4+2$. So, $x_1 = 3.5, x_2 = x_4-1.5$ and $x_1+x_2+x_3+x_4 = 2x_4+2 $ is unbounded.
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
$endgroup$
add a comment |
$begingroup$
No. Let $x_3 = 0$ then, $x_1-x_2 = 5-x_4, x_1+x_2 = x_4+2$. So, $x_1 = 3.5, x_2 = x_4-1.5$ and $x_1+x_2+x_3+x_4 = 2x_4+2 $ is unbounded.
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
$endgroup$
add a comment |
$begingroup$
No. Let $x_3 = 0$ then, $x_1-x_2 = 5-x_4, x_1+x_2 = x_4+2$. So, $x_1 = 3.5, x_2 = x_4-1.5$ and $x_1+x_2+x_3+x_4 = 2x_4+2 $ is unbounded.
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
$endgroup$
No. Let $x_3 = 0$ then, $x_1-x_2 = 5-x_4, x_1+x_2 = x_4+2$. So, $x_1 = 3.5, x_2 = x_4-1.5$ and $x_1+x_2+x_3+x_4 = 2x_4+2 $ is unbounded.
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
answered Jan 9 '17 at 13:26
dezdichadodezdichado
6,4691929
6,4691929
add a comment |
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