A6.Linear Programming PDF

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GALVE, RHEA JEAN S. MATHEMATICS IN MODERN WORLDACTCY11S5-020 ASSIGNMENT 6.Rina needs at least 48 units of protein, 60 units of carbohydrates, and 50 units of fat each month. From each kilogram of food a, she receives 2 units of protein, 4 units of carbohydrates, and 5 units of fats. Food b contains ...

Description

GALVE, RHEA JEAN S.

MATHEMATICS IN MODERN WORLD

ACTCY11S5-020

ASSIGNMENT 6.1

Rina needs at least 48 units of protein, 60 units of carbohydrates, and 50 units of fat each month. From each kilogram of food a, she receives 2 units of protein, 4 units of carbohydrates, and 5 units of fats. Food b contains 3 units of protein, 3 units of carbohydrates, and 2 units of fats. If food A costs Php110 per kilogram and food B costs Php 90 per kilogram. How many kilograms of each food should Sofia buy each month to keep costs at a minimum?

SOLUTION: Let x = kilogram of Food A Let y = kilogram of Food B Let z = minimize costs

Objective Formula: z = Php110x + Php90y

Constraints: Proteins…

2x + 3y ≥ 48

Carbohydrates…

4x + 3y ≥ 60

Fat…

5x + 2y ≥ 50 x, y ≥ 0

Point A 5x + 2y = 50 Let x = 0 5 (0) + 2y = 50 2y 2

=

50 2

y = 25

A is at (0,25)

Point B -3 (5x + 2y = 50) – 3

-4 (5x + 2y = 50) -4

2 (4x + 3y = 60) 2

5 (4x +3y = 60) 5

-15x – 6y = -150 +

-20x – 8y = -200

8x + 6y = 120

+

20x +15y = 300 7y

−7x −30 = −7 −7

7

x = 4.29

=

100 7

y = 14.29

B is at (4.29, 14.29)

Point C -2 (4x + 3y = 60) -2

4x + 3y = 60

2 (2x + 3y = 48) 2

-2 (2x + 3y = 48) -2

-8x – 6y = -120

4x + 3y = 60

+ 4x + 6y = 96

-4x – 6y = - 96

4x 4

=

−24

−3y

4

−3

x=6

=

y = 12

Point D 2x + 3y = 48 Let y = 0 2x + 3(0) = 48 2x 2

=

48 2

x = 24

D is at (24,0)

−36 −3

C is at (6,12)

Vertex

z = Php110x + Php90y

A (0, 25)

Php110(0) + Php90(25) = Php2,250

B (4.29, 14.29)

Php110(4.29) + Php90(14.29) 471.9 + 1,286.1 = Php 1,758

C (6, 12)

Php110(6) + Php90(12) 660 + 1080 =Php 1,740

D (24,0)

Php110(24) + Php90(0) = Php 2,640

Interpretation: Therefore, Rina should buy 6 kilograms of Food A and 12 kilograms of Food B each month to minimize the costs.

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