Answer to Question #173504 in Linear Algebra for ANJU JAYACHANDRAN

Question

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

Answer 1

Solution:

Let the units of product 1 produced be $x$ and the units of product 2 produced be $y$.

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

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

subject to constraints:

$4x+5y\le100 \\ 5x+2y\le80 \\x,y\ge0$

Now, plotting these inequations on graph, we get:

Here, OABC is the feasible region.

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

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

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

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

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

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

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