CUET UG 2025 — Mathematics Calculus

The objective function $z = ax + by$z=ax+by is maximum at two points: $(15, 15)$(15,15) and $(0, 20)$(0,20).

When a linear objective function has maximum value at two different points, both points give the same maximum value. This occurs when the objective function line is parallel to the edge connecting these two points in the feasible region.

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Since both points give the same maximum value:

$$15a + 15b = 0 + 20b$$15a+15b=0+20b

$$15a = 20b - 15b$$15a=20b−15b

$$15a = 5b$$15a=5b

$$3a = b$$3a=b

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Using the condition $ab = 27$ab=27 with $b = 3a$b=3a:

$$a \times 3a = 27$$a×3a=27

$$3a^2 = 27$$3a2=27

$$a^2 = 9$$a2=9

$a = 3$a=3 (since $a \geq 0$a≥0)

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From $b = 3a$b=3a:

$$b = 3 \times 3$$b=3×3

$$b = 9$$b=9

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Using point $(15, 15)$(15,15):

$$z = ax + by$$z=ax+by

$$z = 3(15) + 9(15)$$z=3(15)+9(15)

$$z = 45 + 135$$z=45+135

$$z = 180$$z=180

Therefore, the maximum value of the objective function is $180$180.