Question

A retired person wants to invest an amount of Rs 50,000. His broker recommends investing in the 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 maximise his returns.

Answer 1

Let the retired person invests Rs $x$ in bond A and Rs $y$ in bond B.
Interest on bond A = Rs $x\times \frac{10}{100}=$ Rs $\frac{x}{10}$
Interest on bond B = Rs $y\times \frac{9}{100}=$ Rs $\frac{9y}{100}$
His total interest = Rs $(\frac{x}{10}+\frac{9y}{100})$
Thus, we have linear programming problem (LPP) as to maximize $Z=\frac{x}{10}+\frac{9y}{100}...\left(1\right)$
Subject to constraints:
$x\ge 20,000....\left(2\right)$
$y\ge 10,000....\left(3\right)$
$x\ge y....\left(4\right)$
and $x+y\le 50,\mathrm{000.....}\left(5\right)$
On plotting the inequalities (2) to (5), we have the required region as shown in the graph.
Now, we evaluate the corner points of the shaded region in the graph, These are $A(25,000,25,000),B(20,000,20,000),C(20,000,10,000),D(40,000,10,000)$

$\begin{array}{cc}\text{Corner Point}& Z=\frac{x}{10}+\frac{9y}{100}\\ A(25,000,25,000)& Z=2500+2250=Rs4750\\ B(20,000,20,000)& Z=2000+1800=Rs3800\\ C(20,000,10,000)& Z=2000+900=Rs2900\\ D(40,000,10,000)& Z=4000+900=Rs4900\end{array}$

Here Z is maximum Rs $4900$, when $x=40,000$ and $y=10,000$. Therefore, the retired person should invest Rs $40,000$ in bond A and Rs $10,000$ in bond B.