mathematics-class-12-unit-12-chapter-06-linear-programming-problems-l-6-6-x56wbeypzre

### <Success style="font-size:25px"> Linear Programming Problems L-6 </Success>
### <primary style="font-size:24px"> Problem-1 </primary>
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- A catering agency has two Kitchen to prepare food at two places $A$ and $B$. From these places
'Mid day meal' is to be supplied to three different- schools situated at $P, Q, R$. The monthly  requirements of the school are respectively 40 , 40 and 50 food packets. A packet contains lunch for 100 students. Preparing Capacity of Kitchens $A$ and $B$ are 60 and 70 packets per month respectively. The transportation cost per packet from the Kitchens to school is given below

- Transportation cost per packet (in Rs.)

| To       |    A     |     B      |   
| -------- | -------- | ---------  |   
|     P    |     5    |     4      |   
|     Q    | 4        |     2      |    
|     R    | 3        |     5      |

- How many packets from each kitchens should be transported to schools so that the cost of transportation is minimum? Also, find the minimum cost.

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<footer style = "font-size: 18px"> $\rightarrow$ Linear programming lecture-6 $\rightarrow$ <span style = "color:blue"> Problem-1 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Subject to the constraints,

- x+y $ \leqslant 60 $

- $ x+y \geqslant 10 \quad(x+y-10 \geqslant 0) $

- $  x \leqslant 40 $

- y $ \leqslant 40 $

- x $ \geqslant 0, y \geqslant 0$

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<footer style = "font-size: 18px">Problem-1 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Linear constrants,

- $ x+y \leqslant 60-\text { (i) } $, $ x+y \geqslant 10-\text { (ii) } $

- $ x \leqslant 40-\text { (iii) } $, $ y \leqslant 40-\text { (iv) }$

- Aasociated equation for  (i) , (ii), (iii) \& (iv)

- $ x+y=60 \Rightarrow \frac{x}{60}+\frac{y}{60}=1 $

- x+y =10 $ \Rightarrow \frac{x}{10}+\frac{y}{10}=1 $

- $ x =40$, $y=40 $

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<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-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- By origin test (i)

- 0 + 0 $=0 \leqslant 60$ (True)

- for (i), origin belongs to solution region

- Origin test (ii)

- $0+0=0 \geqslant 10$ (False)

- for (ii), origin does'nt belongs to solution origin

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<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-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ Z_A=3 \times 40+4 \times 20+370=570 $

- $ Z_B=3 \times 20+4 \times 40+370=590 $

- $ Z_C=3 \times 0+4 \times 40+370=530 $

- $ Z_D=3 \times 0+4 \times 10+370=410 $

- $ Z_E=3 \times 10+4 \times 0+370=400 \rightarrow{Min.} $

- $ Z_F=3 \times 40+4 \times 0+370=490$

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

### <Success style="font-size:25px"> Linear Programming Problems L-6 </Success>
### <primary style="font-size:24px"> Problem-2 </primary>
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- The postmaster of a local post office wishes to hire extra helpers during the Deepawali season because of a large increase in volume of mail handling and delivery. Because of the limited office space and budgetary conditions, the number of temporary helpers must not exceed 10. According to past experience, a man can handle 300 letters and 80 packages per day, on the average, and a woman can handle 400 letters and 50 packets per day. The postmaster believes that the daily volume of extra mail and packages will be no less than 3400 and 680 respectively. A man receives 225
- a day and a women receives Rs. 200 a day. How many men and women helpers should be hired to keep the payroll at a minimum? Formulate an LPP and solve it graphically.

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<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-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Let the no. of man hired/day $=x$       
- and the no. of woman hired/day $=y$.         
- Minimise $Z=225 x+200 y$

- Subject to the constraints,

- $ x+y \leq 10 $            
- $ 300 x+400 y \geqslant 3400 \text { i.e. } 3 x+4 y \geqslant 34 $            
- $ 80 x+50 y \geqslant 680 \text { i.e. } 8 x+5 y \geqslant 68 $            
- $ x \geqslant 0, y \geqslant 0 $

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<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-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Associated equation  for (i) , (ii) \& (iii),                              
- $ \frac{x}{10}+\frac{y}{10} $= 1                              
- $ \frac{x }{34 / 3}+\frac{y}{17 / 2}=1 $                              
- $ \frac{x}{17 / 2}+\frac{y}{68 / 5}=1$

- Since all the three lines are concurrent at $P(6,4)$.            
- $\therefore$ feasible region will be Point $P(6,4)$.            
- $\therefore$ The value of Z = $225 \times 6+200 \times 4$ $=2150$            
- $\therefore $ Pay-roll is minimum Rs. $2150 /$ day When 6 man \& 4 woman are employed.         
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Problem-3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Linear Programming Problems L-6 </Success>
### <primary style="font-size:24px"> Problem-3 </primary>
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- The standard weight of a special purpose brick is 5 kg and it must contain two basic ingredients $B_1$ and $B_2$. $B_1$ costs Rs. $5$ per kg and $B_2$ costs Rs. $8$ per kg. Strength consideration dictate that the brick should contain not more than 4 kg of $B_1$ and minimum 2 kg of $\mathrm{B}_2$. Since the demand for the product is likely to be related to the price of the brick, find the minimum cost of brick satisfying the above conditions. Formulate this situation as an LPP and solve it graphically.

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<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-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Let the weight of ingredient $B_1=x$ kg.

- and the weight of ingredient $B_2=y$ kg

| Ingredient | weight(in kg) | cost/kg (in Rs.) |
| --------   | --------      |    --------      |
|   $B_1$    |         x     |      5           |
|   $B_2$    |          y    |      8           |  
|            |     $x+y=5$   |     $Z=5 x+8 y$  |

- x $ \le 4, y \geqslant $ 2

- x $ \geqslant 0, y \geqslant $ 0

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<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-6 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- Corner points

- A(0,5) \& B(3,2)

- $ Z_A=5 \times 0+8 \times 5=40 $        
- $ Z_B=5 \times 3+8 \times 2=31 $        
- $Z_{\text {min }}=31$ at $B(3,2)$        
- $ \therefore$  Weight of $B_1$ = 3 kg           
- Weight of $B_2$ = 2 kg.

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