Linear Programming Word Problem: Theater
I'm having trouble understanding the second constraint.
A theater is presenting a program on drinking and driving for students and their parents[...] admission is 2.00 dollars for parents and 1.00 dollar for students. However, the situation has two constraints: 1) The theater can hold no more than 150 people and 2) every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?
Let $x$ = number of students
$y$ = number of parents
Since the question prompt wants the maximum amount of revenue, then the objective function is $z = x + 2y$
The theater can hold no more than 150 people so the first constraint is simply: $x + y leq 150$
The second constraint is, "every two parents must bring at least one student," but I don't understand how to model this.
I've looked up a solution and it said $y leq 2x$ is how to model this constraint, but I don't understand why. If there must be at least one student for every two parents then why wouldn't the inequality be $2y geq x$?
linear-programming
asked Nov 26 at 1:32
Slecker
1449
add a comment |
I'm having trouble understanding the second constraint.
A theater is presenting a program on drinking and driving for students and their parents[...] admission is 2.00 dollars for parents and 1.00 dollar for students. However, the situation has two constraints: 1) The theater can hold no more than 150 people and 2) every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?
Let $x$ = number of students
$y$ = number of parents
Since the question prompt wants the maximum amount of revenue, then the objective function is $z = x + 2y$
The theater can hold no more than 150 people so the first constraint is simply: $x + y leq 150$
The second constraint is, "every two parents must bring at least one student," but I don't understand how to model this.
I've looked up a solution and it said $y leq 2x$ is how to model this constraint, but I don't understand why. If there must be at least one student for every two parents then why wouldn't the inequality be $2y geq x$?
linear-programming
asked Nov 26 at 1:32
Slecker
1449
add a comment |
I'm having trouble understanding the second constraint.
A theater is presenting a program on drinking and driving for students and their parents[...] admission is 2.00 dollars for parents and 1.00 dollar for students. However, the situation has two constraints: 1) The theater can hold no more than 150 people and 2) every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?
Let $x$ = number of students
$y$ = number of parents
Since the question prompt wants the maximum amount of revenue, then the objective function is $z = x + 2y$
The theater can hold no more than 150 people so the first constraint is simply: $x + y leq 150$
The second constraint is, "every two parents must bring at least one student," but I don't understand how to model this.
I've looked up a solution and it said $y leq 2x$ is how to model this constraint, but I don't understand why. If there must be at least one student for every two parents then why wouldn't the inequality be $2y geq x$?
linear-programming
asked Nov 26 at 1:32
Slecker
1449
I'm having trouble understanding the second constraint.
A theater is presenting a program on drinking and driving for students and their parents[...] admission is 2.00 dollars for parents and 1.00 dollar for students. However, the situation has two constraints: 1) The theater can hold no more than 150 people and 2) every two parents must bring at least one student. How many parents and students should attend to raise the maximum amount of money?
Let $x$ = number of students
$y$ = number of parents
Since the question prompt wants the maximum amount of revenue, then the objective function is $z = x + 2y$
The theater can hold no more than 150 people so the first constraint is simply: $x + y leq 150$
The second constraint is, "every two parents must bring at least one student," but I don't understand how to model this.
I've looked up a solution and it said $y leq 2x$ is how to model this constraint, but I don't understand why. If there must be at least one student for every two parents then why wouldn't the inequality be $2y geq x$?
linear-programming
linear-programming
asked Nov 26 at 1:32
Slecker
1449
asked Nov 26 at 1:32
Slecker
1449
asked Nov 26 at 1:32
Slecker
1449
asked Nov 26 at 1:32
Slecker
1449
asked Nov 26 at 1:32
Slecker
1449
1449
add a comment |
add a comment |
1 Answer
1
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votes
Yes
y <= 2x
The easiest way to test this is with some data points and sketch a graph:
y_______x
2_______1,2,3,4,5…
4_______2,3,4,5,6…
6_______3,4,5,6,7…
Hence:
y <= 2x
2 <= 2x1 True
and
2y >= x
2x2 >= 5 False
Graph of y <= 2x
answered Nov 26 at 3:38
Paul McErlean
13
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Yes
y <= 2x
The easiest way to test this is with some data points and sketch a graph:
y_______x
2_______1,2,3,4,5…
4_______2,3,4,5,6…
6_______3,4,5,6,7…
Hence:
y <= 2x
2 <= 2x1 True
and
2y >= x
2x2 >= 5 False
Graph of y <= 2x
answered Nov 26 at 3:38
Paul McErlean
13
add a comment |
Yes
y <= 2x
The easiest way to test this is with some data points and sketch a graph:
y_______x
2_______1,2,3,4,5…
4_______2,3,4,5,6…
6_______3,4,5,6,7…
Hence:
y <= 2x
2 <= 2x1 True
and
2y >= x
2x2 >= 5 False
Graph of y <= 2x
answered Nov 26 at 3:38
Paul McErlean
13
add a comment |
Yes
y <= 2x
The easiest way to test this is with some data points and sketch a graph:
y_______x
2_______1,2,3,4,5…
4_______2,3,4,5,6…
6_______3,4,5,6,7…
Hence:
y <= 2x
2 <= 2x1 True
and
2y >= x
2x2 >= 5 False
Graph of y <= 2x
answered Nov 26 at 3:38
Paul McErlean
13
Yes
y <= 2x
The easiest way to test this is with some data points and sketch a graph:
y_______x
2_______1,2,3,4,5…
4_______2,3,4,5,6…
6_______3,4,5,6,7…
Hence:
y <= 2x
2 <= 2x1 True
and
2y >= x
2x2 >= 5 False
Graph of y <= 2x
answered Nov 26 at 3:38
Paul McErlean
13
edited Nov 26 at 10:00
answered Nov 26 at 3:38
Paul McErlean
13
answered Nov 26 at 3:38
Paul McErlean
13
answered Nov 26 at 3:38
Paul McErlean
13
13
add a comment |
add a comment |
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