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The corner points of feasible region determined by the following system of linear inequalities 2x+y≤10,x+3y≤15,x≥0,y≥02x + y \leq 10, x + 3y \leq 15, x \geq 0, y \geq 0 are (0,0),(5,0),(3,4)(0,0), (5,0), (3,4) and (0,5)(0,5), then the relation between p and Q so that minimum of Z occurs both points (3,4)(3,4) and (0,5)(0,5) is :

Solution

✅ Correct Option: 2

For Z to be minimum at both (3,4)(3, 4) and (0,5)(0, 5), the values must be equal: 3p+4q=5q3p + 4q = 5q, giving 3p=q3p = q.

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