math 1300: section 5- 3 linear programing in two dimensions: geometric approach

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Linear Programming Problem

Math 1300 Finite MathematicsSection 5.3 Linear Programming In Two Dimensions:

Geometric Approach

Jason Aubrey

Department of MathematicsUniversity of Missouri

Jason Aubrey Math 1300 Finite Mathematics

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Linear programming is a mathematical process that has beendeveloped to help management in decision making and it hasbecome one of the most widely used and best-known tools ofmanagement science.


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In general, a linear programming problem is one that isconcerned with finding the optimal value of a linear objectivefunction of the form

z = ax + by

(where a and b are not both 0) Where the decision variables xand y are subject to the problem constraints.

The problem constraints are various linear inequalities andequations. In all problems we consider, the decision variableswill also satisfy the nonnegative constraints, x ≥ 0, y ≥ 0.


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The set of points satisfying both the problem constraints andthe nonnegative constraints is called the feasible region orfeasible set for the problem.

Any point in the feasible region that produces the optimal valueof the objective function over the feasible region is called anoptimal solution.


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Example: Maximize P = 30x + 40y subject to

2x + y ≤ 10x + y ≤ 7

x + 2y ≤ 12x , y ≥ 0


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Theorem (Fundamental Theorem of Linear Programming)

If the optimal value of the objective function in a linearprogramming problem exists, then that value must occur at one(or more) of the corner points of the feasible region. (A cornerpoint is a point in the feasible set where one or more boundarylines intersect.)


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Applying the theorem...

1 Graph the feasible region, as discussed in Section 5-2. Besure to find all corner points.

2 Construct a corner point table listing the value of theobjective function at each corner point.

3 Determine the optimal solution(s) from the table.4 For an applied problem, interpret the optimal solution(s) in

terms of the original problem.


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Applying the theorem...1 Graph the feasible region, as discussed in Section 5-2. Be

sure to find all corner points.





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sure to find all corner points.2 Construct a corner point table listing the value of the

objective function at each corner point.




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objective function at each corner point.3 Determine the optimal solution(s) from the table.

4 For an applied problem, interpret the optimal solution(s) interms of the original problem.


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objective function at each corner point.3 Determine the optimal solution(s) from the table.4 For an applied problem, interpret the optimal solution(s) in



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Back to our original problem:

Example: Maximize P = 30x + 40y subject to

2x + y ≤ 10x + y ≤ 7

x + 2y ≤ 12x , y ≥ 0


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)


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1 2 3 4 5

1

2

3

4

5

6

0

a

bc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

2x + y ≤ 10Boundary line:

(a) y = 10 − 2xx y0 105 0

Check: 2(0) + 0 ≤︸︷︷︸?

10 Yes!


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1 2 3 4 5

1

2

3

4

5

6

0

ab

c

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

x + y ≤ 7Boundary line:

(b) y = 7 − xx y0 77 0

Check: 0 + 0 ≤︸︷︷︸?

7 Yes!


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

x + 2y ≤ 12Boundary line:

(c) y = 6 − 12x

x y0 6

12 0Check: 0 + 2(0) ≤︸︷︷︸

?

12 Yes!


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

Now we mark the feasible set andfind the coordinates of the cornerpoints.


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

Line c and the y -axis intersect at(0,6)


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

Line c (y = 7 − x) and Line b(y = 6 − 1

2x) intersect at (2,5):

7 − x = 6 − 12

x

−12

x = −1

x = 2y = 7 − 2 = 5


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

Line a (y = 10 − 2x) and Line b(y = 7 − x) intersect at (3,4):

10 − 2x = 7 − x−x = −3

x = 3y = 7 − 3 = 4


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)

(0,0)

Line a intersects the x-axis at(5,0).


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

Don’t forget (0,0)!


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1 2 3 4 5

1

2

3

4

5

6

0

abc

FS

(0,6)

(2,5)

(3,4)

(5,0)(0,0)

Now we construct our corner point ta-ble, and find the maximum value ofthe objective function P = 30x + 40y :Corner Point P = 30x + 40y

(0,6) 240(2,5) 260(3,4) 250(5,0) 150(0,0) 0

Therefore the maximum value of P

occurs at the point (2,5).


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Example: Minimize and maximize P = 20x + 10y subject to

2x + 3y ≥ 302x + y ≤ 26

−2x + 5y ≤ 34x , y ≥ 0


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2 4 6 8 10 12 14

2

4

6

8

10

0

(8,10)

(12,2)

(3,8)

The feasible set has been drawnfor you. We now construct ourcorner point table.Corner Point P = 20x + 10y

(3,8) 140(8,10) 260(12,2) 260

Therefore the maximum value ofP is at (8,10) and (12,2) and theminimum value of P is at (3,8).


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2 4 6 8 10 12 14

2

4

6

8

10

0

(8,10)

(12,2)

(3,8)

The feasible set has been drawnfor you. We now construct ourcorner point table.

Corner Point P = 20x + 10y(3,8) 140(8,10) 260(12,2) 260



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2 4 6 8 10 12 14

2

4

6

8

10

0

(8,10)

(12,2)

(3,8)

The feasible set has been drawnfor you. We now construct ourcorner point table.Corner Point P = 20x + 10y

(3,8) 140(8,10) 260(12,2) 260



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Example: A manufacturing plant makes two types of inflatableboats, a two-person boat and a four-person boat.

Eachtwo-person boat requires 0.9 labor hours from the cuttingdepartment and 0.8 labor-hours from the assembly department.Each four person boat requires 1.8 labor-hours from the cuttingdepartment and 1.2 labor-hours from the assembly department.The maximum labor-hours available per month in the cuttingdepartment and the assembly department are 864 and 672,respectively. The company makes a profit of $25 on each2-person boat and $40 on each four-person boat.

How many of each type should be manufactured each month tomaximize profit? What is the maximum profit?


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Example: A manufacturing plant makes two types of inflatableboats, a two-person boat and a four-person boat. Eachtwo-person boat requires 0.9 labor hours from the cuttingdepartment and 0.8 labor-hours from the assembly department.

Each four person boat requires 1.8 labor-hours from the cuttingdepartment and 1.2 labor-hours from the assembly department.The maximum labor-hours available per month in the cuttingdepartment and the assembly department are 864 and 672,respectively. The company makes a profit of $25 on each2-person boat and $40 on each four-person boat.



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Example: A manufacturing plant makes two types of inflatableboats, a two-person boat and a four-person boat. Eachtwo-person boat requires 0.9 labor hours from the cuttingdepartment and 0.8 labor-hours from the assembly department.Each four person boat requires 1.8 labor-hours from the cuttingdepartment and 1.2 labor-hours from the assembly department.

The maximum labor-hours available per month in the cuttingdepartment and the assembly department are 864 and 672,respectively. The company makes a profit of $25 on each2-person boat and $40 on each four-person boat.



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Example: A manufacturing plant makes two types of inflatableboats, a two-person boat and a four-person boat. Eachtwo-person boat requires 0.9 labor hours from the cuttingdepartment and 0.8 labor-hours from the assembly department.Each four person boat requires 1.8 labor-hours from the cuttingdepartment and 1.2 labor-hours from the assembly department.The maximum labor-hours available per month in the cuttingdepartment and the assembly department are 864 and 672,respectively.

The company makes a profit of $25 on each2-person boat and $40 on each four-person boat.



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Example: A manufacturing plant makes two types of inflatableboats, a two-person boat and a four-person boat. Eachtwo-person boat requires 0.9 labor hours from the cuttingdepartment and 0.8 labor-hours from the assembly department.Each four person boat requires 1.8 labor-hours from the cuttingdepartment and 1.2 labor-hours from the assembly department.The maximum labor-hours available per month in the cuttingdepartment and the assembly department are 864 and 672,respectively. The company makes a profit of $25 on each2-person boat and $40 on each four-person boat.



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To solve such a problem involves:

1 Constructing a mathematical model of the problem, and2 using the mathematical model to find the solution.

So far we have studied some of the techniques we will use forfinding the solution. Here, we introduce a 5 step procedure forconstructing a mathematical model of the problem.

Step 1: Identify the decision variables.

In this problem, we have

x = number of 2-person boats to manufacturey = number of 4-person boats to manufacture


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To solve such a problem involves:1 Constructing a mathematical model of the problem, and

2 using the mathematical model to find the solution.So far we have studied some of the techniques we will use forfinding the solution. Here, we introduce a 5 step procedure forconstructing a mathematical model of the problem.





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To solve such a problem involves:1 Constructing a mathematical model of the problem, and2 using the mathematical model to find the solution.

So far we have studied some of the techniques we will use forfinding the solution. Here, we introduce a 5 step procedure forconstructing a mathematical model of the problem.





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Step 2: Summarize relevant information in table form, relatingthe decision variables with the rows in the table, if possible.

2-person 4-person AvailableCutting 0.9 1.8 864

Assembly 0.8 1.2 672Profit $25 $40

Step 3: Determine the objective and write a linear objectivefunction.

P = 25x + 40y

Step 4: Write problem constraints using linear equationsand/or inequalities.

0.9x + 1.8y ≤ 8640.8x + 1.2y ≤ 672


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Step 2: Summarize relevant information in table form, relatingthe decision variables with the rows in the table, if possible.Recall: Each two-person boat requires 0.9 labor hours from thecutting department and 0.8 labor-hours from the assembly de-partment. Each four person boat requires 1.8 labor-hours fromthe cutting department and 1.2 labor-hours from the assembly de-partment. The maximum labor-hours available per month in thecutting department and the assembly department are 864 and672, respectively. The company makes a profit of $25 on each2-person boat and $40 on each four-person boat.



Step 3: Determine the objective and write a linear objective func-tion.

P = 25x + 40y

Step 4: Write problem constraints using linear equations and/orinequalities.

0.9x + 1.8y ≤ 8640.8x + 1.2y ≤ 672


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P = 25x + 40y

Step 4: Write problem constraints using linear equationsand/or inequalities.

0.9x + 1.8y ≤ 8640.8x + 1.2y ≤ 672


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P = 25x+40y0.9x + 1.8y ≤ 8640.8x + 1.2y ≤ 672

Step 5: Write nonnegative constraints.

x ≥ 0y ≥ 0

We now have a mathematical model of the given problem. Weneed to find the production schedule which results in maximumprofit for the company and to find that maximum profit.


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Now we follow the procedure for geometrically solving a linearprogramming problem with two decision variables.

Applying the Fundamental Theorem of Linear Programming

1 Graph the feasible region, as discussed in Section 5-2. Besure to find all corner points.





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Applying the Fundamental Theorem of Linear Programming1 Graph the feasible region, as discussed in Section 5-2. Be

sure to find all corner points.





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objective function at each corner point.




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objective function at each corner point.3 Determine the optimal solution(s) from the table.

4 For an applied problem, interpret the optimal solution(s) interms of the original problem.


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objective function at each corner point.3 Determine the optimal solution(s) from the table.4 For an applied problem, interpret the optimal solution(s) in



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100 200 300 400 500 600 700 800 900

100

200

300

400

500

(480,240)

FS


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100 200 300 400 500 600 700 800 900

100

200

300

400

500

(480,240)

FS

First we plot the boundaryline 0.9x + 1.8y = 864:

x y0 480

960 0


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100 200 300 400 500 600 700 800 900

100

200

300

400

500

(480,240)

FS

Test:0.9(0) + 1.8(0) ≤︸︷︷︸

?

0 Yes!

So we choose the lower half-

plane.


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100 200 300 400 500 600 700 800 900

100

200

300

400

500

(480,240)

FS

Now plot boundary line0.8x + 1.2y = 672:

x y0 560

840 0


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100 200 300 400 500 600 700 800 900

100

200

300

400

500

(480,240)

FS

Test:0.8(0)+1.2(0) ≤︸︷︷︸

?

672 Yes!

So we choose the lower half-

plane.


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100 200 300 400 500 600 700 800 900

100

200

300

400

500

(480,240)

FS

Next, we find the intersectionpoint of the lines:

480 − 12

x = 560 − 23

x

16

x = 80

x = 480y = 480 − (1/2)(480)= 240


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100 200 300 400 500 600 700 800 900

100

200

300

400

500

(480,240)

FS


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Now we solve the problem:

Corner Point P = 25x + 40y(0,0) 0

(0,480) 19,200(480,240) 21,600(860,0) 21,500

We conclude that the company can make a maximum profit of$21,600 by producing 480 two-person boats and 240four-person boats.


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Remember, there are two parts to solving an applied linearprogramming problem: constructing the mathematical modeland using the geometric method to solve it.


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Example: A chicken farmer can buy a special food mix A at 20cents per pound and a special food mix B at 40 cents perpound. Each pound of mix A contains 3,000 units of nutrient N1and 1,000 units of nutrient N2; each pound of mix B contains4,000 units of nutrient N1 and 4,000 units of nutrient N2. If theminimum daily requirements for the chickens collectively are36,000 units of nutrient N1 and 20,000 units of nutrient N2, howmany pounds of each food mix should be used each day tominimize daily food costs while meeting (or exceeding) theminimum daily nutrient requirements? What is the minimumdaily cost? Construct a mathematical model and solve usingthe geometric method.


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In this problem, we are asked,

how many pounds of each food mix should be usedeach day to minimize daily food costs while meeting(or exceeding) the minimum daily nutrientrequirements?

So,

x = number of pounds of mix A to use each dayy = number of pounds of mix B to use each day


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mix A mix B Min daily req.units of N1 3,000 4,000 36,000units of N2 1,000 4,000 20,000

Cost per pound $0.20 $0.40


We want to minimize daily food costs so the objectivefunction is

C = 0.20x + 0.40y


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Step 2: Summarize relevant information in table form, relatingthe decision variables with the rows in the table, if possible.A chicken farmer can buy a special food mix A at 20 cents perpound and a special food mix B at 40 cents per pound. Eachpound of mix A contains 3,000 units of nutrient N1 and 1,000 unitsof nutrient N2; each pound of mix B contains 4,000 units of nutrientN1 and 4,000 units of nutrient N2. If the minimum daily require-ments for the chickens collectively are 36,000 units of nutrient N1and 20,000 units of nutrient N2. . .



Step 3: Determine the objective and write a linear objective func-tion.

We want to minimize daily food costs so the objective function is

C = 0.20x + 0.40y


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We want to minimize daily food costs so the objectivefunction is

C = 0.20x + 0.40y


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Step 4: Write problem constraints using linear equations and/orinequalities.

3,000x + 4,000y ≥ 36,0001,000x + 4,000y ≥ 20,000

Step 5: Write nonnegative constraints.

x ≥ 0y ≥ 0


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We now have a mathematical model of the given problem:

Minimize C = 0.20x + 0.40ysubject to:

3,000x + 4,000y ≥ 36,0001,000x + 4,000y ≥ 20,000

x ≥ 0y ≥ 0

We will now use the geometric method to determin how manypounds of each food mix should be used each day to minimizedaily food costs while meeting (or exceeding) the minimumdaily nutrient requirements. We can then determine theminimum daily cost as well.


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

And now we plug the corner points into the objective functionto find the minimum cost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

Now plot boundary line 3,000x + 4,000y = 36000:x y0 9

12 0

And now we plug the corner points into the objective func-tion to find the minimum cost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

Test:3,000(0) + 4,000(0) ≥︸︷︷︸

?

36,000 No!

So we choose the upper half-plane.

And now we

plug the corner points into the objective function to find theminimum cost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

Now plot boundary line 1,000x + 4,000y = 20,000:x y0 5

20 0

And now we plug the corner points into the objective func-tion to find the minimum cost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

Test:1,000(0) + 4,000(0) ≥︸︷︷︸

?

20,000 No!

So we choose the upper half-plane.

And now we plug the

corner points into the objective function to find the minimumcost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

Next, we find the intersection point of the lines:

3,000x + 4,000y = 36,000– 1,000x + 4,000y = 20,000

2,000x + 0y = 16,000x = 8

And so, 1,000(8) + 4,000y = 20,000; this implies that y =3.

And now we plug the corner points into the objectivefunction to find the minimum cost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

So, we have the feasible set shown above.

And now weplug the corner points into the objective function to find theminimum cost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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−2 2 4 6 8 10 12 14 16 18 20

2

4

6

8

10

0

(0,9)

(8,3)(20,0)

FS

And now we plug the corner points into the objective functionto find the minimum cost:

Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00


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Point C = 0.20x + 0.40y(0,9) $3.60(8,3) $2.80(20,0) $4.00

We therefore conclude that the chicken farmer can feed thechickens at a minimal cost of $2.80 per day using 8 pounds ofmix A and 3 pounds of mix B.


math 1300: section 5- 3 linear programing in two dimensions: geometric approach

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