CUET UG 2025 — Mathematics Algebra

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.

---

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

---

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.