CUET UG 2025 — Mathematics Calculus

The corner points of the bounded feasible region are: $(5, 5)$(5,5), $(15, 15)$(15,15), $(0, 20)$(0,20), and $(0, 10)$(0,10).

The objective function is $z = 3x + 9y$z=3x+9y.

In Linear Programming, the maximum and minimum values of the objective function occur at the corner points of the feasible region.

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At point $(5, 5)$(5,5):

$$z = 3(5) + 9(5)$$z=3(5)+9(5)

$$z = 15 + 45$$z=15+45

$$z = 60$$z=60

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At point $(15, 15)$(15,15):

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

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

$$z = 180$$z=180

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At point $(0, 20)$(0,20):

$$z = 3(0) + 9(20)$$z=3(0)+9(20)

$$z = 0 + 180$$z=0+180

$$z = 180$$z=180

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At point $(0, 10)$(0,10):

$$z = 3(0) + 9(10)$$z=3(0)+9(10)

$$z = 0 + 90$$z=0+90

$$z = 90$$z=90

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The values of $z$z are: $60, 180, 180, 90$60,180,180,90

Maximum value of $z = 180$z=180

Minimum value of $z = 60$z=60

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$$\text{maximum}(z) - \text{minimum}(z) = 180 - 60$$maximum(z)−minimum(z)=180−60

$$= 120$$=120

Therefore, the value of $\text{maximum}(z) - \text{minimum}(z) = 120$maximum(z)−minimum(z)=120