Let x = belts of A, y = belts of B.
Material constraint: x+y≤800.
Buckles: x≤400, y≤700.
Time: A takes twice the time of B; if B's max is 1000, then 2x+y≤1000.
Objective: Maximize Z = 2x + 1.5y.
Non-negativity: x,y≥0.
A company produces two types of belts A & B with a profit of Rs 2 and Rs 1.50 respectively. Belt of type A needs twice as much time to make as belt…
CUET UG 2023 — Mathematics Applied-Mathematics
2023mcqmedium
A company produces two types of belts A & B with a profit of Rs 2 and Rs 1.50 respectively. Belt of type A needs twice as much time to make as belt type B . The company can produce at the most 1000 belts of type B per day . Material for 800 belts is available per day . At the most , 400 buckles for belt type A and 700 for belts type B are available . Then the appropriate LPP is :
Official previous-year question
Held on 7 Jun 2023 · Verified 13 Jul 2026.
Options
- A
x+y≤800, x≥400, y≥700, 2x+y≤1000, x,y≥0, Max(z) = 2x + 1.5y
- B
x+y≥800, x≥400, y≥700, 2x+y≥1000, x,y≥0, Max(z) = 2x + 1.5y
- C
x+y≤800, x≤400, y≤700, 2x+y≥1000, x,y≥0, Max(z) = 2x + 1.5y
- D
x+y≤800, x≤400, y≤700, 2x+y≤1000, x,y≥0, Max(z) = 2x + 1.5y
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