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 $9 \%$ returns respectively on the invested amounts. He decides to invest atleast 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 $L P P$ graphically to maximise his returns.

Let the person invest Rs x in bond $A$ and Rs $y$ in bond $B$.

Formulation as an LPP.

Maximize $z=0.1 x+0.09 y$ (Returns)

Subject to

$x+y \leqslant 50,000$ (Invertment constraints)

$x \geqslant 20,000$ (investment on bond A constraints.)

$y \geqslant 10,000$ (investment on bond B constraints.)

$x \geqslant y$

$x \geqslant 0, y \geqslant 0$ (non-negative constraints)

Associated equation for these constraints are

$x+y=50,000 \Rightarrow \frac{x}{50000}+\frac{y}{50000}=1$

$x=20,000$

$y=10000$

$x=y$

Origin test for (i)

$0+0=0 \leq 50,000 \text { (true) }$

$\therefore$ origin lies in the solution region for (i)

Arbitrary point test for $x \geqslant y$

Check $(1,0)$

$1 \geqslant 0$ (true)

$\therefore(1,0)$ lies in the solution region

for $x \geqslant y$

for $x \geqslant 20000$

Solution region will be right of the line $x=20,000$

for $y \geqslant 10,000$

Solution region lies above the

line $y=10,000$

After considering all these four constraints

A( 20000,10000)

B (40000, 10000)

C ( 25000, 25000)

D (20000, 20000)

$Z=0.1 x+0.09 y$

$Z_A=0.1 \times 20,000+0.09 \times 10,000=2900$

$Z_B=0.1 \times 40,000+0.09 \times 10,000=4900$

$Z_C=0.1 \times 25000+0.09 \times 25,000=4750$

$Z_D=0.1 \times 20,000+0.09 \times 20,000=3800$

$\therefore Z_{\text {max }}=4900$ at $B(40,000,10,000)$

Hence, the retired person should invest $RS.$ 40,000 in bond $A$ and $Rs. 10,000$ in bond $B$ to get maximum returns $Rs. 4900$ on his investment.

If a young man rides his motorcycle at 25km/hr , he has to spend Rs. 2/km on petrol, if he rides at a faster speed of 40km/hr, the petrol cost increases to Rs. 5/km . He has Rs 100 to spends on petrol and wishes to find the maximum distance he can travel within one hour.

Express this as a LPP and then solve it.

Let x km & y km be the distance covered by the young man at the speed of 25km/hr and 40km/hr resp.

Time consumed in covering these distances are x/25 h and y/40 h resp.

$\therefore$ Total distance travelled

ie. Z = x + y (km.)

Subject to constraints

2x + 5y ⩽ 100

$\begin{gathered}\frac{x}{25}+\frac{y}{50} \leqslant 1 \Rightarrow 8 x+5 y \leqslant 200 , x \geqslant 0, y \geqslant 0\end{gathered}$

linear constraints are

$2 x+5 y \leqslant 100 --(i)$

$8 x+5 y \leqslant 200 --(ii)$

Associated equation for (i) & (ii)

$2 x+5 y=100 \Rightarrow \frac{x}{50}+\frac{y}{20}=1$

$8 x+5 y=200 \Rightarrow \frac{x}{25}+\frac{y}{40}=1$

Origin test for (i)

$2 \times 0+5 \times 0=0 \leqslant 100 \text { (true) }$

$\therefore$ Origin lies in the solution region of (i)

origin test for (ii)

$8 \times 0+5 \times 0=0 \leqslant 200(true)$

$\therefore$ origin lies in the solution region of (ii)

Corner points of the feasible region will be

$A (25,0)$

$B( \frac{50}{3} , \frac{40}{3})$

$C (0,20)$

Let the number of machine $A=x$

Let the number of machine $B=y$

$\therefore$ Maximum Daily Production

$Z=60 x+40 y \text { (Maximize) }$

Subject to constraints

$1000 x+1200 y \leqslant 9000 \text { i.e, } 5 x+6 y \leqslant 45$

$12 x+8 y \leqslant 72 \quad \text { i.e } 3 x+2 y \leqslant 18$

$x \geqslant 0, y \geqslant 0$

Linear constraints are

$5 x+6 y \leqslant 45 \quad \text {- (i) }$

$3 x+2 y \leqslant 18-\text { (ii) }$

Associated equation for (i) and (ii) are

$5 x+6 y=45 \Rightarrow \frac{x}{9}+\frac{y}{7 \cdot 5}=1$

$3 x+2 y=18 \Rightarrow \frac{x}{6}+\frac{y}{9}=1$

Origin test tor (i)

$5 \times 0+6 \times 0=0 \leqslant 45 \text { (true) }$

$\therefore$ origin lies in the solution region of (i)

Origin test for (ii)

$3 \times 0+2 \times 0=0 \leqslant 18 \text { (true)}$

$\therefore$ Origin lies in the solution region of (ii)

Corner points of feasible region are

$A(6,0)$ $B (\frac{9}{4}, \frac{45}{8})$ $C (0 , \frac{15}{2} )$

$Z=60 x+40 y$

$\therefore Z_A=60 \times 6+40 \times 0=360$

$Z_B=60 \times \frac{9}{4}+40 \times \frac{4 5}{8}=135+225=360$

$Z_C=60 \times 0+40 \times \frac{15}{2}=300$

Maximum value of Z lies on line segment $A B$.

No. of machine will be always in integral values.

Hence $Z_{\text {max }}=360$ occurs at$(6,0)$ and $(4,3)$ only.

$\therefore$ Either 6 machine $A$ & no machine $B$

or

4 machine A & 3 machine B will give maximum output.