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Q.Minimise Z=5x+10yZ = 5x + 10y, subject to the constraints x+2y≤120x + 2y \leq 120 x+y≥60x + y \geq 60 x−2y≥0x - 2y \geq 0 x,y≥0x, y \geq 0

CBSECBSE Class XII Board 2025Subjective· 3mImportance★★★★★
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The corner points are A(60,0),B(120,0),C(60,30),D(40,20)A(60,0),B(120,0),C(60,30),D(40,20); evaluating Z=5x+10yZ=5x+10y gives 300,600,600,400300,600,600,400, so the minimum is 300300 at (60,0)(60,0).

Corner-point method: at an optimum, a linear objective Z=ax+byZ=ax+by attains its extreme value at a vertex of the feasible region; compute ZZ at each vertex and compare.

  1. Boundary lines of the constraints: x+2y=120x+2y=120, x+y=60x+y=60, x−2y=0x-2y=0 (i.e. x=2yx=2y), with x≥0, y≥0x\ge0,\ y\ge0.
  2. Find the vertices of the feasible region:
  • A(60,0)A(60,0): on x+y=60x+y=60 and the xx-axis.
  • B(120,0)B(120,0): on x+2y=120x+2y=120 and the xx-axis.
  • C(60,30)C(60,30): intersection of x+2y=120x+2y=120 and x−2y=0x-2y=0 (60+60=12060+60=120, 60−60=060-60=0).
  • D(40,20)D(40,20): intersection of x+y=60x+y=60 and x−2y=0x-2y=0 (40+20=6040+20=60, 40−40=040-40=0). …

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