A retired person wants to invest an amount of Rs. 50,000. His broker recommends investing in two type of bonds A and B yielding 10% and 9% return respectively on the invested amount. He decides to invest at least Rs.20,000 in bond A and at least Rs. 10,000 in bond B. He also wants to invest at least as much in bond A as in bond B. Solve this linear programming problem graphically to maximize his returns.

4900

Points $Z=\frac{10}{100}x+\frac{9}{100}y$

A(20000,10000) Z=Rs.2900

B(40000,10000) Rs.4900   (Maximize)

C(25000,25000) Rs.4750

D(20000,20000) Rs.3800

Let  Rs.x invest in bond A and Rs.y invest in bond B. 

Then A.T.P.  

Maximise, $z=\frac{10}{100}x+\frac{9}{100}y$  .....(1)

Subject to constraints

$x+y≤50,000$ --- (a)

$x≥20,000$---(b)

$y≥10,000$--- (c)

and $x≥y$

or $x−y≥0$--- (d)

and $x≥0,y≥0$

Now change inequality into equations

$x+y=50,000,x=20,000,y=10,000,x=y$

Region: put (0, 0) in (a), (b), (c), (d)

0≤50000 (towards origin)

0≥20,000 (away from origin)

0≥10,000 (away from origin)

As shown in the graph.

So from the tabular column, we conclude that he has to invest Rs.40,000 in 'A' and Rs.10,000 in bond 'B' to get maximum return Rs.4,900.