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NCERT Exemplar · Q5

Q.The feasible region of a linear programming problem is the bounded region in the first quadrant (x≥0x \ge 0, y≥0y \ge 0) satisfying x+2y≤76x + 2y \le 76 and 2x+y≤1042x + y \le 104. Its corner points are O(0,0)O(0, 0), A(52,0)A(52, 0), E(44,16)E(44, 16) and D(0,38)D(0, 38). Determine the maximum value of Z=3x+4yZ = 3x + 4y over this region.

Punjab PsebShort· 3mImportance★★★★★
Appeared in past exams:COMEDK 2024· Set 2024-A· 1mexact
30% · 20/67 Questions
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For a bounded feasible region a linear objective attains its maximum at a corner point. Evaluating Z=3x+4yZ = 3x + 4y at O(0,0)O(0,0), A(52,0)A(52,0), E(44,16)E(44,16), D(0,38)D(0,38) gives 0, 156, 196, 1520,\ 156,\ 196,\ 152, so the maximum value is 196196 at (44,16)(44,16).

Concept

By the Corner Point Theorem, a linear objective function on a bounded convex feasible region attains both its maximum and minimum at a vertex (corner point) of the region. So we only need the value of ZZ at each vertex.

Corner points

The region is { x≥0, y≥0, x+2y≤76, 2x+y≤104 }\{\,x\ge 0,\ y\ge 0,\ x+2y\le 76,\ 2x+y\le 104\,\}. Its vertices are O(0,0)O(0,0), the xx-intercept A(52,0)A(52,0) of 2x+y=1042x+y=104, the yy-intercept D(0,38)D(0,38) of x+2y=76x+2y=76, and the intersection EE of the two boundary lines:

x+2y=76,2x+y=104.x+2y=76,\qquad 2x+y=104. …

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