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Home » A firm manufactures two types of product, A and B, and sells them at a profit of per unit of type and per unit of type B. Each product is processed on two machines, and one unit of type A requires one minute of processing time on and two minutes of processing time on whereas one unit of type requires one minute of processing time on and one minute on Machines and are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximize the profit. Solve the problem graphically.
A firm manufactures two types of product, A and B, and sells them at a profit of \pm 5 per unit of type A and \pm 3 per unit of type B. Each product is processed on two machines, M_{1} and M_{2} . one unit of type A requires one minute of processing time on M_{1} and two minutes of processing time on M_{2} ; whereas one unit of type B requires one minute of processing time on M_{1} and one minute on M_{2} . Machines M_{1} and M_{2} are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximize the profit. Solve the problem graphically.
CBSE | Class 12 | Maths | RS Aggarwal
A firm manufactures two types of product, A and B, and sells them at a profit of \pm 5 per unit of type A and \pm 3 per unit of type B. Each product is processed on two machines, M_{1} and M_{2} . one unit of type A requires one minute of processing time on M_{1} and two minutes of processing time on M_{2} ; whereas one unit of type B requires one minute of processing time on M_{1} and one minute on M_{2} . Machines M_{1} and M_{2} are respectively available for at most 5 hours and 6 hours in a day. Find out how many units of each type of product the firm should produce a day in order to maximize the profit. Solve the problem graphically.

Let the firm manufacture x number of Aand y number of B products.
\therefore According to the question,
X+y \leq 300,2 x+y \leq 360, x \geq 0, y \geq 0
Maximize Z=5 x+3 y
The feasible region determined X+y \leq 300,2 x+y \leq 360, x \geq 0, y \geq 0 is given by

The corner points of feasible region are A(0,0), B(0,300), C(60,240), D(180,0) . The value of Z at corner point is

    \[\begin{tabular}{|l|l|l|} \hline Corner Point & $\mathrm{Z}=5 \mathrm{x}+3 \mathrm{y}$ & \\ \hline $\mathrm{A}(0,0)$ & 0 & \\ \hline $\mathrm{B}(0,300)$ & 900 & \\ \hline $\mathrm{C}(60,240)$ & 1020 & \\ \hline $\mathrm{D}(180,0)$ & 900 & \\ \hline \end{tabular}\]

The maximum value of Z is 1020 and occurs at point (60,240).
The firm should produce 60 Aproducts and 240 B products to earn maximum profit of Rs. 1020 .

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