Mixing
Relationship between sets defined by linear inequalities
$begingroup$
I have the following sets
$$A = {x in [0, 1]^4 mid 83x_1 + 61x_2 + 49x_3 + 20x_4 leq 100}$$
$$B = {x in [0, 1]^4 mid 4x_1 + 3x_2 + 2x_3 + 1x_4 leq 4}$$
How can I prove that $B subset A$?
inequality linear-programming
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
$endgroup$
add a comment |
$begingroup$
I have the following sets
$$A = {x in [0, 1]^4 mid 83x_1 + 61x_2 + 49x_3 + 20x_4 leq 100}$$
$$B = {x in [0, 1]^4 mid 4x_1 + 3x_2 + 2x_3 + 1x_4 leq 4}$$
How can I prove that $B subset A$?
inequality linear-programming
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
$endgroup$
5
$begingroup$
Multiply the second inequality by 20.
$endgroup$
– Jean Marie
Jan 28 at 19:41
1
$begingroup$
By proving $4x_1 + 3x_2 + 2x_3 +x_4 le 4 implies 83x_1 + 61x_2 + 49x_3 + 20x_4 le 100$ for $0le x_i le 1$.
$endgroup$
– fleablood
Jan 28 at 19:50
add a comment |
$begingroup$
I have the following sets
$$A = {x in [0, 1]^4 mid 83x_1 + 61x_2 + 49x_3 + 20x_4 leq 100}$$
$$B = {x in [0, 1]^4 mid 4x_1 + 3x_2 + 2x_3 + 1x_4 leq 4}$$
How can I prove that $B subset A$?
inequality linear-programming
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
$endgroup$
I have the following sets
$$A = {x in [0, 1]^4 mid 83x_1 + 61x_2 + 49x_3 + 20x_4 leq 100}$$
$$B = {x in [0, 1]^4 mid 4x_1 + 3x_2 + 2x_3 + 1x_4 leq 4}$$
How can I prove that $B subset A$?
inequality linear-programming
inequality linear-programming
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
edited Jan 31 at 7:52
Rodrigo de Azevedo
13.2k41960
13.2k41960
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
asked Jan 28 at 19:38
J. KuhnJ. Kuhn
61
61
5
$begingroup$
Multiply the second inequality by 20.
$endgroup$
– Jean Marie
Jan 28 at 19:41
1
$begingroup$
By proving $4x_1 + 3x_2 + 2x_3 +x_4 le 4 implies 83x_1 + 61x_2 + 49x_3 + 20x_4 le 100$ for $0le x_i le 1$.
$endgroup$
– fleablood
Jan 28 at 19:50
add a comment |
5
$begingroup$
Multiply the second inequality by 20.
$endgroup$
– Jean Marie
Jan 28 at 19:41
1
$begingroup$
By proving $4x_1 + 3x_2 + 2x_3 +x_4 le 4 implies 83x_1 + 61x_2 + 49x_3 + 20x_4 le 100$ for $0le x_i le 1$.
$endgroup$
– fleablood
Jan 28 at 19:50
5
5
$begingroup$
Multiply the second inequality by 20.
$endgroup$
– Jean Marie
Jan 28 at 19:41
$begingroup$
Multiply the second inequality by 20.
$endgroup$
– Jean Marie
Jan 28 at 19:41
1
1
$begingroup$
By proving $4x_1 + 3x_2 + 2x_3 +x_4 le 4 implies 83x_1 + 61x_2 + 49x_3 + 20x_4 le 100$ for $0le x_i le 1$.
$endgroup$
– fleablood
Jan 28 at 19:50
$begingroup$
By proving $4x_1 + 3x_2 + 2x_3 +x_4 le 4 implies 83x_1 + 61x_2 + 49x_3 + 20x_4 le 100$ for $0le x_i le 1$.
$endgroup$
– fleablood
Jan 28 at 19:50
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $x_ige0$. Multiply second inequation by $25$. Then
$$100x_1+75x_2+50x_3+25x_1le100,$$
$$83x_1 + 61x_2 + 49x_3 + 20x_4le 100x_1+75x_2+50x_3+25x_1le100$$
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Let $x_ige0$. Multiply second inequation by $25$. Then
$$100x_1+75x_2+50x_3+25x_1le100,$$
$$83x_1 + 61x_2 + 49x_3 + 20x_4le 100x_1+75x_2+50x_3+25x_1le100$$
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
Let $x_ige0$. Multiply second inequation by $25$. Then
$$100x_1+75x_2+50x_3+25x_1le100,$$
$$83x_1 + 61x_2 + 49x_3 + 20x_4le 100x_1+75x_2+50x_3+25x_1le100$$
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
add a comment |
$begingroup$
Let $x_ige0$. Multiply second inequation by $25$. Then
$$100x_1+75x_2+50x_3+25x_1le100,$$
$$83x_1 + 61x_2 + 49x_3 + 20x_4le 100x_1+75x_2+50x_3+25x_1le100$$
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
$endgroup$
Let $x_ige0$. Multiply second inequation by $25$. Then
$$100x_1+75x_2+50x_3+25x_1le100,$$
$$83x_1 + 61x_2 + 49x_3 + 20x_4le 100x_1+75x_2+50x_3+25x_1le100$$
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
answered Jan 28 at 21:09
Aleksas DomarkasAleksas Domarkas
1,62317
1,62317
add a comment |
add a comment |
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5
$begingroup$
Multiply the second inequality by 20.
$endgroup$
– Jean Marie
Jan 28 at 19:41
1
$begingroup$
By proving $4x_1 + 3x_2 + 2x_3 +x_4 le 4 implies 83x_1 + 61x_2 + 49x_3 + 20x_4 le 100$ for $0le x_i le 1$.
$endgroup$
– fleablood
Jan 28 at 19:50