Question #151319

2A firm manufactures two products A and B on which the profits earned per unit are Rs. 3 and Rs.4 respectively. Each product is processed on two machines M and M Product Arequires one minute of processing time on M, and two minutes on M, while requires one minute on M, and one minute on M, Machine M is available for not more than 7 hours 30 minutes while machine M is available for 10 hours during any working day. Find the number of units of products A and B to be manufactured to get maximum profit. Formulate the above as a LPP and solve by graphical method.


1
Expert's answer
2020-12-17T07:10:11-0500

Let xx units of firm A and yy units of firm B were manufactured: x0,y0.x\geq0, y\geq0.

Given information can be tabulated as


Time of M1Time of M2Product A (x)1min2minProduct B (y)1min1minAvailability450min600min\def\arraystretch{1.5} \begin{array}{c:c:c} & Time\ of\ M_1 & Time\ of\ M_2 \\ \hline Product\ A\ (x) & 1\min & 2\min \\ Product\ B\ (y) & 1\min & 1\min \\ \hdashline Availability & 450\min & 600\min \end{array}

The constrants are


x+y450,2x+y600x+y\leq450, 2x+y\leq600

Total profit Z=3x+4yZ=3x+4y which is to be maximized.

The objective function: Max Z=3x+4yMax\ Z=3x+4y

Subjective Constraints

x+y450x+y\leq450

x+2y600x+2y\leq600

x0,y0x\geq0, y\geq0

Horizontal (x)(x)axis: Product A , Vertical (y)(y) axis: Product B

Constraint No. 1: x+y450x+y\leq450 Converting into equality: 

x+y=450x+y=450

The two points which make the constraint line are: (0,450)(0,450) and (450,0)(450, 0)


Constraint No. 2: x+2y600x+2y\leq600 Converting into equality: 

x+2y=600x+2y=600

The two points which make the constraint line are: (0,300)(0,300) and (600,0)(600, 0)

Each constraint will be represented by a single straight line on the graph. There are two constraints, hence there will be two straight lines.

The feasible region determined by the system of constraints


x+y450x+2y600x0y0\begin{matrix} x+y\leq450 \\ x+2y\leq600\\ x\geq 0 \\ y\geq0 \end{matrix}

is as follows



The corner points are O(0,0),D(0,300),E(300,150),A(450,0).O(0,0), D(0,300),E(300,150),A(450,0).

The values of ZZ at these corner points are as follows


Corner pointZ=3x+4yO0D1200E1500A1350\begin{matrix} Corner\ point & Z=3x+4y \\ O & 0 \\ D & 1200\\ E & 1500\\ A & 1350 \end{matrix}

The maximum value of ZZ is Rs1500Rs1500 which is attained at E(300,150).E(300,150).


The maximum profit is Rs1500Rs1500 obtained when 300 units of firm A and 150 products of firm B are manufactured.



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