The corner points of the bounded feasible region determined by a set of constraints in an LPP are P(0, 5), Q(3, 5), R(5, 0) and S(4, 1). If the…

The Linear Programming Problem has corner points P(0, 5), Q(3, 5), R(5, 0), and S(4, 1) with objective function $z = ax + 2by$z=ax+2by where $a, b > 0$a,b>0.

When the maximum occurs at two corner points, both points must give the same z-value, and this value must be greater than or equal to the z-values at other corner points.

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At P(0, 5):

$$z = a(0) + 2b(5)$$z=a(0)+2b(5)

$$z = 10b$$z=10b

At Q(3, 5):

$$z = a(3) + 2b(5)$$z=a(3)+2b(5)

$$z = 3a + 10b$$z=3a+10b

At R(5, 0):

$$z = a(5) + 2b(0)$$z=a(5)+2b(0)

$$z = 5a$$z=5a

At S(4, 1):

$$z = a(4) + 2b(1)$$z=a(4)+2b(1)

$$z = 4a + 2b$$z=4a+2b

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Since maximum occurs at both Q and S, their z-values must be equal:

$$3a + 10b = 4a + 2b$$3a+10b=4a+2b

$$10b - 2b = 4a - 3a$$10b−2b=4a−3a

$$8b = a$$8b=a

$$a - 8b = 0$$a−8b=0

Therefore, the condition is $a - 8b = 0$a−8b=0.