Let x,y denote pieces of M1,M2. Profit (to maximize): Z=800x+1200y. Fabricating constraint: 9x+12y≤180, simplifies to 3x+4y≤60. Finishing constraint: x+3y≤30. Non-negativity: x,y≥0.
CUET UG 2023 — Mathematics Algebra
A manufacturing company makes two models M1 and M2 of a product. Each piece of M1 requires 9 labour hours for fabricating and one labour hour for finishing. Each piece of M2 require 12 labour hours for fabricating and 3 labour hours for finishing. For fabricating and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of Rs.800 on each piece of M1 and Rs.1200 on each piece of M2
The above Linear Programming Problem [LPP] is given by
Held on 23 May 2023 · Verified 13 Jul 2026.
Maximize Z = 800x + 1200y
Subject to constraints,
3x+4y≤60
x+3y≤30
x,y≥0
Maximize Z = 800x + 1200y
Subject to constraints,
3x+4y≥60
x+3y≥30
x,y≥0
Minimize Z = 800x + 1200y
Subject to constraints,
3x+4y≤60
x+3y≥30
x,y≥0
Minimize Z = 800x + 1200y
Subject to constraints,
3x+4y≥60
x+3y≤30
x,y≥0
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