A manufacturer has three machines $M_1, M_2$ and $M_3$ installed in his factory. Machines $M_1$ and $M_2$ are capable of being operated for atmost 12 hours, whereas machine $M_3$ must be operated for atleast 5 hours a day. The manufacturer produces only two items, each requiring the use of these three machines. The following table gives the number of hours required on these machines for producing 1 unit of A or B.

He makes a profit of ₹60 on item A and ₹40 on item B. He wishes to find out how many of each item he should produce to have maximum profit. Formulate it as L.P.P.

Maximize $Z=60x+40y$
Subject to:
$x+2y≤12$
$2x+y≤12$
$x + \frac{5}{4}y≥5$
$x≥0,y≥0$

The correct answer is Option (2) → Maximize $Z=60x+40y$, Subject to: $x+2y≤12, 2x+y≤12, x + \frac{5}{4}y≥5, x≥0,y≥0$

Let the manufacturer produce x units of item A and y units of item B. As he makes a profit of 60 on one item of A and 40 on one item of B, his total profit (in) is

$Z = 60x + 40y$.

From the given data, we can mathematically formulate the L.P.P. as

Maximize $Z = 60x + 40y$ subject to the constraints

$x + 2y ≤ 12$ (Machine $M_1$ constraint)

$2x + y ≤ 12$ (Machine $M_2$ constraint)

$x +\frac{5}{4} y ≥ 5$ (Machine $M_3$ constraint)

$x≥0, y ≥0$ (Non-negativity constraints)