The corner points of the bounded feasible region determined by the system of linear constraints are (0, 0), (5, 0), (3, 4) and (0, 5). Let Z = px +…

The corner points of the bounded feasible region are (0, 0), (5, 0), (3, 4) and (0, 5). For the objective function $Z = px + qy$Z=px+qy where $p, q > 0$p,q>0, the maximum occurs at two corner points when both points give equal Z values.

For maximum at both (5, 0) and (3, 4), these points must yield the same value of Z.

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At point (5, 0):

$$Z = p(5) + q(0)$$Z=p(5)+q(0)

$$Z = 5p$$Z=5p

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At point (3, 4):

$$Z = p(3) + q(4)$$Z=p(3)+q(4)

$$Z = 3p + 4q$$Z=3p+4q

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For maximum to occur at both points:

$$5p = 3p + 4q$$5p=3p+4q

$$5p - 3p = 4q$$5p−3p=4q

$$2p = 4q$$2p=4q

$$p = 2q$$p=2q

This can also be written as $2q = p$2q=p.

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In linear programming, when the maximum occurs at two corner points, the objective function line is parallel to the edge connecting those points. The condition $2q = p$2q=p ensures this property.

Therefore, the condition is $2q = p$2q=p.