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)?

Max $Z = 10000 x + 15000 y$, subject to following constraints $3x + 4y ≤ 60, x + 3y ≤ 30, x≥0,y≥0$

The correct answer is Option (4) → Max $Z = 10000 x + 15000 y$, subject to following constraints $3x + 4y ≤ 60, x + 3y ≤ 30, x≥0,y≥0$

Let $x$ be the number of classic models and $y$ be the number of supreme models

Profit function

$Z=10000x+15000y$

Assembling time constraint

$9x+12y\le180$

Finishing time constraint

$x+3y\le30$

Non-negativity conditions

$x\ge0,\;y\ge0$

The required LPP is:

Maximize $Z=10000x+15000y$ subject to $9x+12y\le180,\;x+3y\le30,\;x\ge0,\;y\ge0$.