mathematics-class-12-unit-12-chapter-04-linear-programming-problems-l-4-6-otdhtd2csuw

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
<hr>
<main>

- 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.

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px"> $\rightarrow$ Linear programming lecture-4 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

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

| Type of bond | no. of bonds | Returns on bonds |
| -------- | -------- | -------- |
|  A   | x    | 10%     |
| B    | y     | 9%     |
|     | x+y⩽ 50,000     | z = 10% of x + 9% of y    |
|     | x ⩾ 0 ,y ⩾ 0    |  = 0.1x + 0.09y       |

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Linear programming lecture-4 $\rightarrow$ Problem-1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-2 </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
<hr>
<main>

-  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)

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-2 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- 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 $

</div>
<div style='float: right; width:1%;'>

</div>
<div style='float: right; width:30%;'>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/cb9002d2-20ff-4363-ba7c-e793f4e14700/public)

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- 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 $x=20,000$

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- A( 20000,10000)

- B (40000, 10000)

- C ( 25000, 25000)

- D (20000, 20000)

</div>
<div style='float: right; width:1%;'>

</div>
<div style='float: right; width:30%;'>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fd0b6294-05f1-444f-4785-7f575befc900/public)

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-3 </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
<hr>
<main>

-  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.

</div>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- 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

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

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-3 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- 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 $

</div>
<div style='float: right; width:1%;'>

</div>
<div style='float: right; width:30%;'>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/ad047f36-58b8-4208-ed95-9c8e0204af00/public)

</main>
<hr>
<footer style = "font-size: 18px">Problem-3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- Corner points of the feasible region will be

- $ A (25,0) $

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

- $ C (0,20)$

</div>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/df40c1fa-b08e-4cdf-7c59-4e2acf03de00/public)

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Problem-4 </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Problem-4 </primary>
<hr>
<main>

-  A factory owner purchases two types of machines $A$ and $B$, for his factory. The requirement  and limitations for the machines are as follows

| Machine | Area occupied by the machine | Labour force by each machine | Daily output (in units) |
| -------- | -------- | -------- | -------- |
|  A    | 1000 sq.m   | 12 man   | 60   |
| B    | 1200 sq.m     | 8 man | 40   |

- He has an area of 9000 Sq.m available and 72 skilled men who can operate the machines. How many machine of each type should he buy to maximize the daily output?

</div>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Problem-4 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- 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 $

</div>
<div style='float: right; width:1%;'>

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Problem-4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- 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$

</div>

</div>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d66c12a5-1b14-4e31-0982-ca490ebc4d00/public)

</main>
<hr>
<footer style = "font-size: 18px">Problem-4 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- 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} ) $

</div>
<div style='float: right; width:1%;'>

</div>
<div style='float: right; width:25%;'>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/b776ab51-ee0a-4b66-0806-9202d91f7600/public)

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-4 </Success>
### <primary style="font-size:24px"> Solution </primary>
<hr>
<main>

- $ 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.

</div>

![alt text](https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/11215312-e69d-414f-9c40-29dddb32f200/public)

</main>
<hr>
<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>