Let x and y be the number of beg A and B 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+12x2
Subject to ,40x1+30x2≥150
20x1+20x2≥90
10x1+30x2≥60,x1≥0,x2≥0.
(B). Use graphical approach to find the solution
1. To draw constraint 40x1+30x2≥150→(1)Treat it as 40x1+30x2=150(RED LINE IN GRAPH)x1 0 3.75x2 5 02. To draw constraint 20x1+20x2≥90→(2)Treat it as 20x1+20x2=90(GREEN LINE IN GRAPH)x1 0 4.5x2 4.5 03. To draw constraint 10x1+30x2≥60→(3)Treat it as 10x1+30x2=60(BLACK LINE IN GRAPH)x1 0 6x2 2 0

The value of the objective function at each of these extreme points is as follows:
Extreme Point Objective function value
Coordinates Z=10x1+12x2
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 : x1 =3.75, x2 =0.75 and min Z=46.5.
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