How to formula the given linear programming model?
Chem Labs uses raw materials I and II to produce two domestic cleaning solutions, A and B.
The daily availabilities of raw materials I and II are 150 and 145 units, respectively. One unit of
solution A consumes 0.5 unit of raw material I and 0.6 unit of raw material II, and one unit of
solution B uses 0.5 unit of raw material I and 0.4 unit of raw material II. The profits per unit of
solutions A and B are 8 and $10, respectively. The daily demand for solution A lies between
30 and 150 units, and that for solution B between 40 and 200 units. Find the optimal production
amounts of A and B.
My attempt
Let A and B be the no. of units of A and B produced and X and Y be no. of raw materials I and II to be processed respectively.
The objective function is to maximize the profit, Z.
Z=8A+10B
The objective function is subject to the following constraints
30<=0.5X+0.6Y<=150
40<=0.5X+0.4Y<=200
X<=150
Y<=145
Is this formulation correct? If it is, how can one proceed from this point to find the maximum profit?
linear-programming
add a comment |
Chem Labs uses raw materials I and II to produce two domestic cleaning solutions, A and B.
The daily availabilities of raw materials I and II are 150 and 145 units, respectively. One unit of
solution A consumes 0.5 unit of raw material I and 0.6 unit of raw material II, and one unit of
solution B uses 0.5 unit of raw material I and 0.4 unit of raw material II. The profits per unit of
solutions A and B are 8 and $10, respectively. The daily demand for solution A lies between
30 and 150 units, and that for solution B between 40 and 200 units. Find the optimal production
amounts of A and B.
My attempt
Let A and B be the no. of units of A and B produced and X and Y be no. of raw materials I and II to be processed respectively.
The objective function is to maximize the profit, Z.
Z=8A+10B
The objective function is subject to the following constraints
30<=0.5X+0.6Y<=150
40<=0.5X+0.4Y<=200
X<=150
Y<=145
Is this formulation correct? If it is, how can one proceed from this point to find the maximum profit?
linear-programming
add a comment |
Chem Labs uses raw materials I and II to produce two domestic cleaning solutions, A and B.
The daily availabilities of raw materials I and II are 150 and 145 units, respectively. One unit of
solution A consumes 0.5 unit of raw material I and 0.6 unit of raw material II, and one unit of
solution B uses 0.5 unit of raw material I and 0.4 unit of raw material II. The profits per unit of
solutions A and B are 8 and $10, respectively. The daily demand for solution A lies between
30 and 150 units, and that for solution B between 40 and 200 units. Find the optimal production
amounts of A and B.
My attempt
Let A and B be the no. of units of A and B produced and X and Y be no. of raw materials I and II to be processed respectively.
The objective function is to maximize the profit, Z.
Z=8A+10B
The objective function is subject to the following constraints
30<=0.5X+0.6Y<=150
40<=0.5X+0.4Y<=200
X<=150
Y<=145
Is this formulation correct? If it is, how can one proceed from this point to find the maximum profit?
linear-programming
Chem Labs uses raw materials I and II to produce two domestic cleaning solutions, A and B.
The daily availabilities of raw materials I and II are 150 and 145 units, respectively. One unit of
solution A consumes 0.5 unit of raw material I and 0.6 unit of raw material II, and one unit of
solution B uses 0.5 unit of raw material I and 0.4 unit of raw material II. The profits per unit of
solutions A and B are 8 and $10, respectively. The daily demand for solution A lies between
30 and 150 units, and that for solution B between 40 and 200 units. Find the optimal production
amounts of A and B.
My attempt
Let A and B be the no. of units of A and B produced and X and Y be no. of raw materials I and II to be processed respectively.
The objective function is to maximize the profit, Z.
Z=8A+10B
The objective function is subject to the following constraints
30<=0.5X+0.6Y<=150
40<=0.5X+0.4Y<=200
X<=150
Y<=145
Is this formulation correct? If it is, how can one proceed from this point to find the maximum profit?
linear-programming
linear-programming
asked Dec 29 '18 at 2:56
HamadaHamada
82
82
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2 Answers
2
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oldest
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The use of raw materials $I$ and $II$ depends on the production of $A$ and $B$, so you don't need the variables $X$ and $Y$.
You can create a table of given data:
$$begin{array}{c|c|c|c}
Products&I&II&Profit&Demand\
hline
A&0.5&0.6&8&30le Ale 150\
B&0.5&0.4&10&40le Ble 200\
hline
Available&le 150&le 145&maximize&end{array}$$
Now we can formulate the LPP: let $A$ and $B$ be the numbers of units of $A$ and $B$, respectively. Then:
$$pi(A,B)=8A+10Bto text{max} text{subject to}\
0.5A+0.5Ble 150 text{(material I constraint)}\
0.6A+0.4Ble 145 text{(material II constraint)}\
30le Ale 150 text{(demand for A)}\
40le Ble 200 text{(demand for B)}\
$$
You can use graphical or Simplex methods to solve LPP.
Graphical method.
1) Draw the feasible (green) region from the constraint inequalities:
2) Find the corner points: $A,B,C,D,E,F$.
3) Evaluate the objective (profit) function at the corner points and choose the maximum.
Can you do it?
Answer:
$pi(100,200)=2800.$ WolframAlpha answer.
add a comment |
We just have to decide how many units of $A$ and $B$ are to be produced.
You are right that the profit is $8A+10B$ and we want to maximize it.
Now, let's examine the constraint imposed by material I.
$$0.5A+0.5B le 150$$
Now, let's examine the constraint imposed by material II.
$$0.6A + 0.4B le 145$$
The demand informations also gives us
$$30 le A le 150$$
and
$$40 le Ble 200.$$
Now, we have a $2$-dimensional linear programming problem, you can use a graphical method to solve the problem if you wish.
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The use of raw materials $I$ and $II$ depends on the production of $A$ and $B$, so you don't need the variables $X$ and $Y$.
You can create a table of given data:
$$begin{array}{c|c|c|c}
Products&I&II&Profit&Demand\
hline
A&0.5&0.6&8&30le Ale 150\
B&0.5&0.4&10&40le Ble 200\
hline
Available&le 150&le 145&maximize&end{array}$$
Now we can formulate the LPP: let $A$ and $B$ be the numbers of units of $A$ and $B$, respectively. Then:
$$pi(A,B)=8A+10Bto text{max} text{subject to}\
0.5A+0.5Ble 150 text{(material I constraint)}\
0.6A+0.4Ble 145 text{(material II constraint)}\
30le Ale 150 text{(demand for A)}\
40le Ble 200 text{(demand for B)}\
$$
You can use graphical or Simplex methods to solve LPP.
Graphical method.
1) Draw the feasible (green) region from the constraint inequalities:
2) Find the corner points: $A,B,C,D,E,F$.
3) Evaluate the objective (profit) function at the corner points and choose the maximum.
Can you do it?
Answer:
$pi(100,200)=2800.$ WolframAlpha answer.
add a comment |
The use of raw materials $I$ and $II$ depends on the production of $A$ and $B$, so you don't need the variables $X$ and $Y$.
You can create a table of given data:
$$begin{array}{c|c|c|c}
Products&I&II&Profit&Demand\
hline
A&0.5&0.6&8&30le Ale 150\
B&0.5&0.4&10&40le Ble 200\
hline
Available&le 150&le 145&maximize&end{array}$$
Now we can formulate the LPP: let $A$ and $B$ be the numbers of units of $A$ and $B$, respectively. Then:
$$pi(A,B)=8A+10Bto text{max} text{subject to}\
0.5A+0.5Ble 150 text{(material I constraint)}\
0.6A+0.4Ble 145 text{(material II constraint)}\
30le Ale 150 text{(demand for A)}\
40le Ble 200 text{(demand for B)}\
$$
You can use graphical or Simplex methods to solve LPP.
Graphical method.
1) Draw the feasible (green) region from the constraint inequalities:
2) Find the corner points: $A,B,C,D,E,F$.
3) Evaluate the objective (profit) function at the corner points and choose the maximum.
Can you do it?
Answer:
$pi(100,200)=2800.$ WolframAlpha answer.
add a comment |
The use of raw materials $I$ and $II$ depends on the production of $A$ and $B$, so you don't need the variables $X$ and $Y$.
You can create a table of given data:
$$begin{array}{c|c|c|c}
Products&I&II&Profit&Demand\
hline
A&0.5&0.6&8&30le Ale 150\
B&0.5&0.4&10&40le Ble 200\
hline
Available&le 150&le 145&maximize&end{array}$$
Now we can formulate the LPP: let $A$ and $B$ be the numbers of units of $A$ and $B$, respectively. Then:
$$pi(A,B)=8A+10Bto text{max} text{subject to}\
0.5A+0.5Ble 150 text{(material I constraint)}\
0.6A+0.4Ble 145 text{(material II constraint)}\
30le Ale 150 text{(demand for A)}\
40le Ble 200 text{(demand for B)}\
$$
You can use graphical or Simplex methods to solve LPP.
Graphical method.
1) Draw the feasible (green) region from the constraint inequalities:
2) Find the corner points: $A,B,C,D,E,F$.
3) Evaluate the objective (profit) function at the corner points and choose the maximum.
Can you do it?
Answer:
$pi(100,200)=2800.$ WolframAlpha answer.
The use of raw materials $I$ and $II$ depends on the production of $A$ and $B$, so you don't need the variables $X$ and $Y$.
You can create a table of given data:
$$begin{array}{c|c|c|c}
Products&I&II&Profit&Demand\
hline
A&0.5&0.6&8&30le Ale 150\
B&0.5&0.4&10&40le Ble 200\
hline
Available&le 150&le 145&maximize&end{array}$$
Now we can formulate the LPP: let $A$ and $B$ be the numbers of units of $A$ and $B$, respectively. Then:
$$pi(A,B)=8A+10Bto text{max} text{subject to}\
0.5A+0.5Ble 150 text{(material I constraint)}\
0.6A+0.4Ble 145 text{(material II constraint)}\
30le Ale 150 text{(demand for A)}\
40le Ble 200 text{(demand for B)}\
$$
You can use graphical or Simplex methods to solve LPP.
Graphical method.
1) Draw the feasible (green) region from the constraint inequalities:
2) Find the corner points: $A,B,C,D,E,F$.
3) Evaluate the objective (profit) function at the corner points and choose the maximum.
Can you do it?
Answer:
$pi(100,200)=2800.$ WolframAlpha answer.
answered Dec 29 '18 at 7:44
farruhotafarruhota
19.6k2737
19.6k2737
add a comment |
add a comment |
We just have to decide how many units of $A$ and $B$ are to be produced.
You are right that the profit is $8A+10B$ and we want to maximize it.
Now, let's examine the constraint imposed by material I.
$$0.5A+0.5B le 150$$
Now, let's examine the constraint imposed by material II.
$$0.6A + 0.4B le 145$$
The demand informations also gives us
$$30 le A le 150$$
and
$$40 le Ble 200.$$
Now, we have a $2$-dimensional linear programming problem, you can use a graphical method to solve the problem if you wish.
add a comment |
We just have to decide how many units of $A$ and $B$ are to be produced.
You are right that the profit is $8A+10B$ and we want to maximize it.
Now, let's examine the constraint imposed by material I.
$$0.5A+0.5B le 150$$
Now, let's examine the constraint imposed by material II.
$$0.6A + 0.4B le 145$$
The demand informations also gives us
$$30 le A le 150$$
and
$$40 le Ble 200.$$
Now, we have a $2$-dimensional linear programming problem, you can use a graphical method to solve the problem if you wish.
add a comment |
We just have to decide how many units of $A$ and $B$ are to be produced.
You are right that the profit is $8A+10B$ and we want to maximize it.
Now, let's examine the constraint imposed by material I.
$$0.5A+0.5B le 150$$
Now, let's examine the constraint imposed by material II.
$$0.6A + 0.4B le 145$$
The demand informations also gives us
$$30 le A le 150$$
and
$$40 le Ble 200.$$
Now, we have a $2$-dimensional linear programming problem, you can use a graphical method to solve the problem if you wish.
We just have to decide how many units of $A$ and $B$ are to be produced.
You are right that the profit is $8A+10B$ and we want to maximize it.
Now, let's examine the constraint imposed by material I.
$$0.5A+0.5B le 150$$
Now, let's examine the constraint imposed by material II.
$$0.6A + 0.4B le 145$$
The demand informations also gives us
$$30 le A le 150$$
and
$$40 le Ble 200.$$
Now, we have a $2$-dimensional linear programming problem, you can use a graphical method to solve the problem if you wish.
answered Dec 29 '18 at 5:11
Siong Thye GohSiong Thye Goh
99.9k1465117
99.9k1465117
add a comment |
add a comment |
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