Answer to Question #222591 in Operations Research for HANNA

Question #222591

Problem 1: A farmer plans to mix two types of food to make a mix of low cost feed

for the animals in his farm. A bag of food A costs $10 and contains 40 units of

proteins, 20 units of minerals and 10 units of vitamins. A bag of food B costs $12

and contains 30 units of proteins, 20 units of minerals and 30 units of vitamins. How

many bags of food A and B should the consumed by the animals each day in order to

meet the minimum daily requirements of 150 units of proteins, 90 units of minerals

and 60 units of vitamins at a minimum cost?

A. Formulate the problem in to linear programming problem model

B. Use graphical approach to find the solution


1
Expert's answer
2021-08-03T12:56:19-0400

Let x and yx \ \text{and} \ y be the number of beg AA and BB respectively to meet the minimum daily requirements of food.

Then according to question,


(A)'Formulate the problem in to linear programming problem model'


Minimize ,Z=10x1+12x2Z=10x_1+12x_2

Subject to ,40x1+30x215040x_1+30x_2\ge150

20x1+20x29020x_1+20x_2\geq90

10x1+30x260,x10,x20.10x_1+30x_2\ge60,x_1\geq0,x_2\ge0.

(B). Use graphical approach to find the solution

1. To draw constraint 40x1+30x2150(1)Treat it as 40x1+30x2=150(RED LINE IN GRAPH)x1      0      3.75x2      5      02. To draw constraint 20x1+20x290(2)Treat it as 20x1+20x2=90(GREEN LINE IN GRAPH)x1      0      4.5x2      4.5      03. To draw constraint 10x1+30x260(3)Treat it as 10x1+30x2=60(BLACK LINE IN GRAPH)x1      0      6x2      2      01. \space To \space draw \space constraint \space 40x_1+30x_2≥150→(1) \\ \\ Treat \space it \space as \space 40x_1+30x_2=150 \\ (RED \space LINE \space IN \space GRAPH) \\ \\ x_1 \space \space \space \space \space \space 0 \space \space \space \space \space \space 3.75 \\ x_2 \space \space \space \space \space \space 5 \space \space \space \space \space \space 0 \\ \\ 2. \space To \space draw \space constraint \space 20x_1+20x_2≥90→(2) \\ \\ Treat \space it \space as \space 20x_1+20x_2=90 \\ (GREEN \space LINE \space IN \space GRAPH) \\ \\ x_1 \space \space \space \space \space \space 0 \space \space \space \space \space \space 4.5 \\ x_2 \space \space \space \space \space \space 4.5 \space \space \space \space \space \space 0 \\ \\ 3. \space To \space draw \space constraint \space 10x_1+30x_2≥60→(3) \\ \\ Treat \space it \space as \space 10x_1+30x_2=60 \\ (BLACK \space LINE \space IN \space GRAPH) \\ \\ x_1 \space \space \space \space \space \space 0 \space \space \space \space \space \space 6 \\ x_2 \space \space \space \space \space \space 2 \space \space \space \space \space \space 0 \\





The value of the objective function at each of these extreme points is as follows:


Extreme Point Objective function value

Coordinates Z=10x1+12x2Z=10x_1+12x_2

A(0,5) 10(0)+12(5)=60

B(1.5,3) 10(1.5)+12(3)=51

C(3.75,0.75) 10(3.75)+12(0.75)=46.5

D(6,0) 10(6)+12(0)=60



The minimum value of the objective function Z=46.5 occurs at the extreme point (3.75,0.75).


Hence, the optimal solution to the given LP problem is : x1x_1 =3.75, x2x_2 =0.75 and min Z=46.5.






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