# Linear Programming

 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}

 A Region P B Region Q C Region R D Region S
GATE ME 2023   Industrial Engineering
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, $Z_{max}=3900\; for \; (x_1,x_2)=(20,50)$
 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 __________.
$\begin{array}{|c|c|c|c|c|} \hline &\text{M1 } &\text{M2}&\text{M3}&\text{Profit per piece (Rs.)} \\ \hline P1 & 2 &2 &0&150\\ \hline P2 & 3& 1&2&100\\ \hline \text{Maximum quantity available per day}& 70& 50&40\\ \hline \end{array}$
 A 5000 B 4000 C 3000 D 6000
GATE ME 2022 SET-2   Industrial Engineering
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}

 Question 3
In a linear programming problem, if a resource is not fully utilized, the shadow price of that resource is
 A positive B negative C zero D infinity
GATE ME 2022 SET-1   Industrial Engineering
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)
 A 6 B 7 C 12 D 10
GATE ME 2020 SET-1   Industrial Engineering
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
 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 ____________.
 A 12 B 15 C 18 D 20
GATE ME 2017 SET-1   Industrial Engineering
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

There are 5 questions to complete.

### 5 thoughts on “Linear Programming”

1. 6th question must give < 8. in question it has given 0.