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Question
Solve the following Linear Programming Problem graphically:
Minimize: Z = 60x + 80y
Subject to constraints:
3x + 4y ≥ 8
5x + 2y ≥ 11
x, y ≥ 0
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Solution
Minimize Z = 60x + 80y
Subject to constraints
3x + 4y ≥ 8
5x + 2y ≥ 11
x, y ≥ 0
Change inequation to equations, then we get
3x + 4y = 8 ...(i)
5x + 2y = 11 ...(ii)
`A(8/3, 0), Z = 60 xx 8/3 + 80 xx 0` = 160
`B(2, 1/2), Z = 60 xx 2 + 80 xx 1/2` = 160
`C(0, 11/2),Z = 60 xx 0 + 80 xx 11/2` = 440
As feasible region is unbounded, therefore 160 may or may not be minimum value. So we will have to graph the inequality 60x + 80y < 160 or 3x + 4y < 8. It can be seen that feasible region has no common points with 3x + 4y < 8. So the minimum cost of mixture will be ₹160 at the line segment joining `(8/3, 0)` and `(2, 1/2)`.
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