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Linear Programming Problem Solving

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by

Maddie Uilk

on 11 November 2012

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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
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