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# Linear Programming Problem Solving

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by

Tweet## Maddie Uilk

on 11 November 2012#### Transcript of Linear Programming Problem Solving

A farmer has 90 acres available for planting millet and alfalfa. Seeds cost $4 per acre for millet and $6 per acre for alfalfa. Labor costs are $20 per acre for millet and $10 per acre for alfalfa. The expected income is $110 per acre for millet and $150 per acre of alfalfa. The farmer intends to spend no more than $480 for seed and $1400 for labor. Find the income for the given constraints. Using Linear Programming to Solve Real Life Problems

By: Maddie Uilk Constraints:

4x + 6y ≤ 480

20x + 10y ≤ 1400

x + y ≤ 90

x ≥0

y ≥ 0 Define Variables:

x = millet

y = alfalfa Step 2: Step 1: Step 3: Find the feasible region: Simplified Constraints:

y ≤ -2/3x + 80

y ≤ -2x + 140

y ≤ -x + 90

x ≥ 0

y ≥ 0 Step 4: A= (0,0)

B= (0, 80)

C= (30,60)

D= (50,40)

E= (20,0) A B C D E Step 5: Profit Statement:

P= 110x + 150y P= 110x + 150y

A (0,0) =$0

B (0,80) =$12,000

C (30, 60) = $12,300

D (50,40) = $11,500

E (20,0) = $7,700 1)

2)

3)

4)

5) 1)

2)

3)

4)

5) Step 6: Profit Statement:

In order to maximize income of $12,300 the farmer needs to sell 30 millet acres and 60 alfalfa acres. Find Vertices: Need to Find C:

4x + 6y ≤ 480

x+ y ≤ 90 4x + 6y ≤ 480

4(x + y ≤ 90) 4x + 6y ≤ 480

4x + 4y ≤ 360

____________

2y ≤ 120

____________

2

Y≤ 60 - 4x + 6(60) ≤ 480 4x + 360 ≤ 480 - 360 360 _______________ 4x ≤ 120

_______

4

X≤ 30 Need to Find D:

20x + 10y ≤ 1400

x + y ≤ 90 20x + 10y ≤ 1400

10(x + y ≤ 90) 20x + 10y ≤ 1400

10x + 10y ≤ 900

_______________

10x ≤ 500

__________

10

X≤ 50 - x(50) + y ≤ 90

50 + y ≤ 90

50 50

____________

Y≤ 40 - A: P=110(0) + 150(0) = 0

B: P=110(0) + 150(80) = 12,000

C: P=110(30) + 150(60) = 12,300

D: P=110(50) + 150(40)= 11,500

E: P=110(20) + 150(0) = 7,750 4x + 6y ≤ 480

-4x -4x

______________

6y ≤ -4x + 480

______________

6

Y ≤ -2/3x + 80 20x + 10y ≤ 1400

-20x 20x

________________

10y ≤ -20x + 1400

_________________

10

Y≤ -2x + 140 x + y ≤ 90

-x -x

___________

Y ≤ -x + 90

Full transcriptBy: Maddie Uilk Constraints:

4x + 6y ≤ 480

20x + 10y ≤ 1400

x + y ≤ 90

x ≥0

y ≥ 0 Define Variables:

x = millet

y = alfalfa Step 2: Step 1: Step 3: Find the feasible region: Simplified Constraints:

y ≤ -2/3x + 80

y ≤ -2x + 140

y ≤ -x + 90

x ≥ 0

y ≥ 0 Step 4: A= (0,0)

B= (0, 80)

C= (30,60)

D= (50,40)

E= (20,0) A B C D E Step 5: Profit Statement:

P= 110x + 150y P= 110x + 150y

A (0,0) =$0

B (0,80) =$12,000

C (30, 60) = $12,300

D (50,40) = $11,500

E (20,0) = $7,700 1)

2)

3)

4)

5) 1)

2)

3)

4)

5) Step 6: Profit Statement:

In order to maximize income of $12,300 the farmer needs to sell 30 millet acres and 60 alfalfa acres. Find Vertices: Need to Find C:

4x + 6y ≤ 480

x+ y ≤ 90 4x + 6y ≤ 480

4(x + y ≤ 90) 4x + 6y ≤ 480

4x + 4y ≤ 360

____________

2y ≤ 120

____________

2

Y≤ 60 - 4x + 6(60) ≤ 480 4x + 360 ≤ 480 - 360 360 _______________ 4x ≤ 120

_______

4

X≤ 30 Need to Find D:

20x + 10y ≤ 1400

x + y ≤ 90 20x + 10y ≤ 1400

10(x + y ≤ 90) 20x + 10y ≤ 1400

10x + 10y ≤ 900

_______________

10x ≤ 500

__________

10

X≤ 50 - x(50) + y ≤ 90

50 + y ≤ 90

50 50

____________

Y≤ 40 - A: P=110(0) + 150(0) = 0

B: P=110(0) + 150(80) = 12,000

C: P=110(30) + 150(60) = 12,300

D: P=110(50) + 150(40)= 11,500

E: P=110(20) + 150(0) = 7,750 4x + 6y ≤ 480

-4x -4x

______________

6y ≤ -4x + 480

______________

6

Y ≤ -2/3x + 80 20x + 10y ≤ 1400

-20x 20x

________________

10y ≤ -20x + 1400

_________________

10

Y≤ -2x + 140 x + y ≤ 90

-x -x

___________

Y ≤ -x + 90