A manufacturing unit makes two models, 'classic' and 'supreme' of the scooter. Each piece of the classic model requires 9 labour hours for assembling and 1 labour hour for finishing. Each piece of supreme model requires 12 labour hours for assembling and 3 labour hour for finishing. For assembling and finishing, the maximum labour hours available are 180 and 30 respectively. The company makes a profit of ₹ 10000 on each piece of the classic model and ₹ 15000 on each piece of the supreme model. Which of the following options describes the given Linear Programming Problem (LPP) to maximize the profit Z (Max Z) (where x and y are the number of pieces of the classic model and the supreme model respectively)?

Let $x$ = number of Classic scooters and $y$ = number of Supreme scooters.

Labour hours needed per scooter:

• Classic: 9 hours assembling, 1 hour finishing
• Supreme: 12 hours assembling, 3 hours finishing

Available resources:

• Maximum assembling hours = 180
• Maximum finishing hours = 30

Profit per scooter:

• Classic = ₹10,000
• Supreme = ₹15,000

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The profit function to maximize:

Max $Z = 10000x + 15000y$

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Total assembling hours used cannot exceed 180 hours:

$9x + 12y \leq 180$

Dividing by 3:

$3x + 4y \leq 60$

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Total finishing hours used cannot exceed 30 hours:

$x + 3y \leq 30$

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Non-negativity constraints:

$x \geq 0, y \geq 0$

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The complete LPP formulation:

Max $Z = 10000x + 15000y$

Subject to:

• $3x + 4y \leq 60$ (assembling constraint)
• $x + 3y \leq 30$ (finishing constraint)
• $x \geq 0, y \geq 0$ (non-negativity)

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Checking the options:

Option 1: Wrong profit function ($10x + 15y$) and wrong constraint ($3x + 4y \leq 180$)

Option 2: Wrong constraints ($9x + y \leq 180$ and $12x + 3y \leq 30$)

Option 3: Wrong inequality signs ($\geq$ instead of $\leq$)

Option 4: Correct formulation with $Z = 10000x + 15000y$ subject to $3x + 4y \leq 60, x + 3y \leq 30, x \geq 0, y \geq 0$

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Therefore, the answer is Option 4.