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Q.Maximise z=300x+190yz = 300x + 190y subject to constraints : x+y≤24,x + y \leq 24, 2x+y≤32,2x + y \leq 32, x≥0,y≥0.x \geq 0, y \geq 0.

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★
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Max z=300x+190y=5440z=300x+190y=5440 at (8,16)(8,16).

For a bounded LPP the optimum of a linear objective occurs at a corner (vertex) of the feasible region — evaluate zz at each vertex.

  1. Constraints: x+y≤24, 2x+y≤32, x≥0, y≥0.x+y\le 24,\ 2x+y\le 32,\ x\ge 0,\ y\ge 0.
  2. Corner points of the feasible region:
    • (0,0)(0,0);
    • (16,0)(16,0) from 2x+y=32, y=02x+y=32,\ y=0;
    • (0,24)(0,24) from x+y=24, x=0x+y=24,\ x=0;
    • intersection of x+y=24x+y=24 and 2x+y=322x+y=32: subtracting gives x=8, y=16x=8,\ y=16, i.e. (8,16)(8,16).
  3. Evaluate z=300x+190yz=300x+190y: …

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