Question 1 |

Which one of the options given represents the feasible region of the linear
programming model:

\begin{aligned} Maximize\;\; 45X_1&+60X_2 \\ X_1&\leq 45 \\ X_2&\leq 50 \\ 10X_1+10X_2& \geq 600 \\ 25X_1+5X_2&\leq 750 \end{aligned}

\begin{aligned} Maximize\;\; 45X_1&+60X_2 \\ X_1&\leq 45 \\ X_2&\leq 50 \\ 10X_1+10X_2& \geq 600 \\ 25X_1+5X_2&\leq 750 \end{aligned}

Region P | |

Region Q | |

Region R | |

Region S |

Question 1 Explanation:

\begin{aligned}
x_1&=45 &...(i)\\
x_2&= 50&...(ii)\\
10x_1+10x_2&=600 \\
or\; x_1+x_2&=60&...(iii) \\
25x_1+5x_2&=750 \\
or\; 5x_1+x_2&=150&...(iv) \\
\end{aligned}

By drawing the curve we get 3 values of x_1 and x_2 as (10, 50), (20, 50), (22.5, 37.5)

So, Z_{max}=45x_1+60x_2 for (10,50)

Z_{max}=450+3000=4450

for (20,50)

Z_{max}=45\times 20+50 \times 60=3900

for (22.5, 37.5)

Z_{max}=45\times 22.5+60 \times 37.5=3262.5

So, Z_{max}=3900\; for \; (x_1,x_2)=(20,50)

By drawing the curve we get 3 values of x_1 and x_2 as (10, 50), (20, 50), (22.5, 37.5)

So, Z_{max}=45x_1+60x_2 for (10,50)

Z_{max}=450+3000=4450

for (20,50)

Z_{max}=45\times 20+50 \times 60=3900

for (22.5, 37.5)

Z_{max}=45\times 22.5+60 \times 37.5=3262.5

So, Z_{max}=3900\; for \; (x_1,x_2)=(20,50)

Question 2 |

A manufacturing unit produces two products P1
and P2. For each piece of P1 an P2, the table below
provides quantities of materials M1, M2, and M3
required, and also the profit earned. The maximum
quantity available per day for M1, M2 and M3 is
also provided. The maximum possible profit per
day is Rs __________.

\begin{array}{|c|c|c|c|c|} \hline &\text{M1 } &\text{M2}&\text{M3}&\text{Profit per piece (Rs.)} \\ \hline P1 & 2 &2 &0&150\\ \hline P2 & 3& 1&2&100\\ \hline \text{Maximum quantity available per day}& 70& 50&40\\ \hline \end{array}

\begin{array}{|c|c|c|c|c|} \hline &\text{M1 } &\text{M2}&\text{M3}&\text{Profit per piece (Rs.)} \\ \hline P1 & 2 &2 &0&150\\ \hline P2 & 3& 1&2&100\\ \hline \text{Maximum quantity available per day}& 70& 50&40\\ \hline \end{array}

5000 | |

4000 | |

3000 | |

6000 |

Question 2 Explanation:

Let,

x_1= No. of units of product P_1

x_2= No. of units of product P_2

Max, Z=150x_1+100x_2

\begin{aligned} s.t.\;\; 2x_1+3x_2 &\leq 70 \\ 2x_1+3x_2&\leq 50 \\ 2x_2 &\leq 40 \\ x_1\geq 0, x_2&\geq 0 \end{aligned}

Point B: Intersection of

\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_2 &\leq 40 \\ x_2=20, x_1&=5 \end{aligned}

Point C: Intersection of

\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_1+x_2 &\leq 50 \\ \text{By sloving }2x_2 &=20 \\ \Rightarrow x_2=10, x_1&=20 \end{aligned}

\begin{aligned} Z_{max}&=150x_1+100x_2\\ Z|_{A(0,20)}&=150(0)+100(20)=2000\\ Z|_{B(5,20)}&=150(5)+100(20)=2750\\ Z|_{C(20,10)}&=150(20)+100(10)=4000\\ Z|_{D(25,0)}&=150(25)+100(0)=3750\\ \Rightarrow x_1&=20,x_2=10\\ Z_{max}&=4000 \end{aligned}

x_1= No. of units of product P_1

x_2= No. of units of product P_2

Max, Z=150x_1+100x_2

\begin{aligned} s.t.\;\; 2x_1+3x_2 &\leq 70 \\ 2x_1+3x_2&\leq 50 \\ 2x_2 &\leq 40 \\ x_1\geq 0, x_2&\geq 0 \end{aligned}

Point B: Intersection of

\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_2 &\leq 40 \\ x_2=20, x_1&=5 \end{aligned}

Point C: Intersection of

\begin{aligned} 2x_1+3x_2 &\leq 70 \\ 2x_1+x_2 &\leq 50 \\ \text{By sloving }2x_2 &=20 \\ \Rightarrow x_2=10, x_1&=20 \end{aligned}

\begin{aligned} Z_{max}&=150x_1+100x_2\\ Z|_{A(0,20)}&=150(0)+100(20)=2000\\ Z|_{B(5,20)}&=150(5)+100(20)=2750\\ Z|_{C(20,10)}&=150(20)+100(10)=4000\\ Z|_{D(25,0)}&=150(25)+100(0)=3750\\ \Rightarrow x_1&=20,x_2=10\\ Z_{max}&=4000 \end{aligned}

Question 3 |

In a linear programming problem, if a resource is
not fully utilized, the shadow price of that resource
is

positive | |

negative | |

zero | |

infinity |

Question 3 Explanation:

In a Linear programming problem if a resource is
not fully utilized. The shadow price or dual price
of the particular resource is zero.

Question 4 |

Two business owners Shveta and Ashok run their businesses in two different states. Each
of them, independent of the other, produces two products A and B, sells them at Rs.
2,000 per kg and Rs, 3,000 per kg, respectively, and uses Linear Programming to
determine the optimal quantity of A and B to maximize their respective daily revenue.
Their constraints are as follows: i) for each business owner, the production process is
such that the daily production of A has to be at least as much as B, and the upper
limit for production of B is 10 kg per day, and ii) the respective state regulations restrict
Shveta?s production of A to less than 20 kg per day and Ashok's production of A to
less than 15 kg per day. The demand of both A and B in both the states is very high
and everything produced is sold.

The absolute value of the difference in daily (optimal) revenue of Shveta and Ashok is ________ thousand Rupees (round off to 2 decimal places)

The absolute value of the difference in daily (optimal) revenue of Shveta and Ashok is ________ thousand Rupees (round off to 2 decimal places)

6 | |

7 | |

12 | |

10 |

Question 4 Explanation:

\begin{aligned} \text { Maximum } z &=2000 x_{1}+3000 x_{2} \\ A \rightarrow x_{1} \text { units } \quad x_{1} &\geq x_{2} \\ B \rightarrow x_{2} \text { units } \quad x_{2} &\geq 10 \\ x_{1} & \lt 20 \\ x_{1} & \lt 15 \\ \text { Shveta's Profit } &=\text { Rs. } 70000 \text { at }(20,10) \\ \text { Ashok's Profit } &=\text { Rs. } 60000 \text { at }(15,10) \end{aligned}

Difference Rs. 10000

Difference Rs. 10000

Question 5 |

Two models, P and Q, of a product earn profits of Rs. 100 and Rs. 80 per piece, respectively. Production times for P and Q are 5 hours and 3 hours, respectively, while the total production time available is 150 hours. For a total batch size of 40, to maximize profit, the number of units of P to be produced is ____________.

12 | |

15 | |

18 | |

20 |

Question 5 Explanation:

Max. Z=100 x P + 80 x Q

\begin{aligned} 5P+3Q&\leq 150\\ P+Q&\leq 40\\ \frac{P}{30}+\frac{Q}{30}&=1 \end{aligned}

Putting value of corner point in objective function

Z(A)=3200

Z(B)=3500

Z(C)=3000

Maximum at B,

P=15, Q=25

\begin{aligned} 5P+3Q&\leq 150\\ P+Q&\leq 40\\ \frac{P}{30}+\frac{Q}{30}&=1 \end{aligned}

Putting value of corner point in objective function

Z(A)=3200

Z(B)=3500

Z(C)=3000

Maximum at B,

P=15, Q=25

There are 5 questions to complete.

6th question must give < 8. in question it has given 0.

Thank you for your suggestions. We have updated the correction suggested by You.

in question no 30…i think Z is max at x2 = 10

pardon….question no 20

QN-11, ANS IS B

SIMPLEX METHOD USES ONLY CORNER POINTS OF THE FEASIBLE REGION AS ONE OF THE CORNER POINTS HAS MAXIMUM VALUE.