For Z to be minimum at both (3,4) and (0,5), the values must be equal: 3p+4q=5q, giving 3p=q.
The corner points of feasible region determined by the following system of linear inequalities 2x+y≤10,x+3y≤15,x≥0,y≥0 are (0,0),(5,0),(3,4) and (0,5), then the relation between p and Q so that minimum of Z occurs both points (3,4) and (0,5) is :
Held on 11 Aug 2022 · Verified 13 Jul 2026.
p=3q
3p=q
3q+2p
2p=2(q−2)
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