A firm manufactures two products \(A\) and \(B\) on which profit earned per unit are ₹3 and ₹4 respectively. Each product is processed on two machines \(M_1\) and \(M_2\). The product A requires one minute of processing time on \(M_1\) and two minute…

Answer: Make 0 units of A and 450 units of B; maximum profit ₹1800.

Let \(x\) units of A and \(y\) units of B be made. Maximise \(Z=3x+4y\) subject to

\(x+y\le450\) (\(M_1\)), \(2x+y\le600\) (\(M_2\)), \(x\ge0,\ y\ge0\).

Line \(x+y=450\) meets the axes at \((450,0)\), \((0,450)\); line \(2x+y=600\) meets them at \((300,0)\), \((0,600)\). Both inequalities are satisfied by the origin, so the feasible region lies on the origin side of both lines, in the first quadrant.

Solving \(x+y=450\) and \(2x+y=600\): \(x=150\), \(y=300\).

[Rough graph: feasible region OABC with O(0,0), A(300,0), B(150,300), C(0,450).]

Maximum \(Z=1800\) at \(C(0,450)\). The firm should make 0 units of A and 450 units of B for a maximum profit of ₹1800.