Question #173504

8. (a) A manufacturer has two products P1

 and , P2

 both of which are produced in two steps 

by machines M1

 and . M2

 The process time per hundred for the products on the 

machines

M1 M2

Profit (in 

thousand Rs. 

per 100 units) 

P1

4 5 10 

P2

5 2 5 

Available 

hours 

100 80 

 The manufacturer can sell as much as he can produce of both products. Formulate the 

problem as LP model. Determine optimum solution, u


1
Expert's answer
2021-04-15T06:39:15-0400

Solution:

Let the units of product 1 produced be xx and the units of product 2 produced be yy.

Rs.1000 per 100 units, or Rs. 10 per 1 unit.

Objective function, maximise Z=10(10x)+10(5y)=100x+50yZ=10(10x)+10(5y)=100x+50y

subject to constraints:

4x+5y1005x+2y80x,y04x+5y\le100 \\ 5x+2y\le80 \\x,y\ge0

Now, plotting these inequations on graph, we get:



Here, OABC is the feasible region.

Corner points are O(0,0),A(16,0),B(20017,18017),C(0,20)O(0,0), A(16,0),B(\frac{200}{17},\frac{180}{17}),C(0,20)

At O(0,0),Z=100(0)+50(0)=0O(0,0), Z=100(0)+50(0)=0

At A(16,0),Z=100(16)+50(0)=1600A(16,0), Z=100(16)+50(0)=1600

At B(20017,18017),Z=100(20017)+50(18017)6470.58B(\frac{200}{17},\frac{180}{17}), Z=100(\frac{200}{17})+50(\frac{180}{17})\approx6470.58

At O(0,20),Z=100(0)+50(20)=1000O(0,20), Z=100(0)+50(20)=1000

Clearly, maximum occurs at B(20017,18017)B(\frac{200}{17},\frac{180}{17}).

Thus, the optimal solution is Rs. 6470.58 when 20017\frac{200}{17}units of product 1 and 18017\frac{180}{17} units of product 2 are produced.


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