Linear Programmig Problem MCQ Quiz in বাংলা - Objective Question with Answer for Linear Programmig Problem - বিনামূল্যে ডাউনলোড করুন [PDF]

Calculation:

Converting the inequations to equations, we obtain:

2x + y = 3, x + 2y = 6, x = 0, y = 0.

For 2x + y = 3 

• Putting x = 0, 0 + y = 3 ⇒ y = 3
• Putting y = 0, 2x + 0 = 3 ⇒ x = \(\frac{3}{2}\)
• This line meets the x-axis at \(\left ( \frac{3}{2},0 \right )\) and the y-axis at (0, 3). Draw a line through these points.
• We see that the origin (0, 0) does not satisfy the inequation 2x + y > 3
• Therefore, the region that does not contain the origin is the solution of the inequality 2x + y > 3

For x + 2y = 6

• Putting x = 0, 0 + 2y = 6 ⇒ y = 3
• Putting y = 0, x + 0 = 6 ⇒ x = 6
• This line meets the x-axis at (6, 0) and the y-axis at (0, 3). Draw a line through these points.
• We see that the origin (0, 0) does not satisfy the inequation x + 2y > 6
• Therefore, the region that does not contain the origin is the solution of the inequality x + 2y > 6

The solution to the inequalities is the intersection of the 2x + y > 3, x + 2y > 6 and x, y > 0. Thus, the shaded region represents the solution set of the given set of inequalities.

Now solving 2x + y = 3 and x + 2y = 6, we get the point P(0, 3).

Now, Putting (3, 5) in 2x + y > 3,

2(3) + 5 > 3 ⇒ 11 > 3 which is true.

Again, putting (3, 5) in x + 2y > 6,

3 + 2(5) > 6 ⇒ 13 > 6 which is true.

∴ (3, 5) lies in the common region.

So, the correct option is (4)

Solution:

Given that, Z = 11x + 7y and the constraints 2x + y ≤ 6, x ≤ 2, x ≥ 0, y ≥ 0

Let 2x + y = 6

Now, plotting all the constrain equations we see that the shaded area OABC is the feasible region determined by the constraints.

The feasible region is bounded. So, the maximum value will occur at a corner point of the feasible region.

Corner points are (0, 0), (2, 0), (2, 2) and (0, 6).

On evaluating the value of Z, we get

CONCEPT:

• At the corner point of the feasible reason, we have to check the value of the objective function to get the maximum value.

CALCULATION:

Given:

So, the condition of p and q, so that the minimum of Z occurs at (3, 0) and (1, 1) can be extracted by equating the minimum values of Z for (3, 0) and (1, 1).

⇒ p + q = 3p

⇒ 2p = q

∴ \(\rm p=\frac{q}{2}\)

• So, the correct answer will be option 2.

Explanation:

 Z = px + qy, where p, q > 0

Corner points of the feasible region determined by the system of linear constraints are (0, 3), (1, 1), and (3, 0)

Let z₀​ be the minimum value of z in the feasible region.

The value z_0​ is attained at both (3, 0) and (1, 1).

z_0​ = p × 3 + q × 0     ---- (1)

z₀ ​= p × 1 + q × ​1     ---- (2)

From Equations (1) and (2) we get 

⟹ p × 3 + q × 0 = p × 1 + q × ​1 

⟹ 3 p = p + q

⟹ 2 p = q

⟹ p = q/2

The Correct Option is (2).

Solution:

Given that, Z = 5x + 3y and the constraints 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0

Taking 3x + 5y = 15, we have

And, taking 5x + 2y = 10, we have

Now, plotting all the constrain equations we see that the shaded area OABC is the feasible region determined by the constraints.

The feasible region is bounded. So, the maximum value will occur at a corner point of the feasible region.

Corner points are (0, 0), (2, 0), (0, 3) and \(\left ( \frac{20}{19},\frac{45}{19} \right )\)

On evaluating the value of Z, we get

Concept: 

There are three main components of linear programming:

• Decision Variables: These variables are the activities that share the available resources while also competing with one another. In other words, these are interrelated in terms of resource utilization and need to be solved simultaneously. They are assumed to be non-negative and continuous.  
• The Objective Function: Each linear programming problem is aimed to have an objective to be measured in quantitative terms such as profits, sales, cost, time, or some other parameter, etc which needs to be maximized or minimized. The objective function will vary depending upon the business requirements or other factors.
• Constraints: These represent real-life limitations such as money, time, labor or other restrictions and the objective function needs to be maximized or minimized while satisfying the constraints. They are represented as linear equations or inequations in terms of the decision variables.

For example, suppose x and y are the decision variables. The objective function will be given by:

Z = ax + by ….(1)

Where a and b are constants and Z is the function to be maximized or minimized. There will be conditions x ≥ 0, and y ≥ 0, which indicates the non-negative constraints on the decision variable.

The equation looks very simple since there are various assumptions involved while forming a linear programming example. These are mentioned below:

• Parameters such as resources available, profit contribution of unit decision variable and resource used by unit decision variable needs to be known.
• Decision variables are continuous. Hence the outputs can be an integer or a fraction.
• The contribution of each decision variable in the objective function is directly proportional to the objective function.

Calculation:

Given:

x, y: Number of units of A and B being sold respectively.

It is very beneficial if the information is converted to a tabular form to make the data visualization much easier. As per the above info, the table is formed below:

• Since x and y cannot be negative, x ≥ 0      …(2) and y ≥ 0      ….(3)
• One unit of A takes 1 unit of electricity and 3 units of diesel.   
• Similarly, one unit of B takes 1 unit of electricity and 2 units of diesel.   

⇒ x + y ≤ 5       ….(4) and 3x + 2y ≤ 12       ….(5)

• The objective function will be the profit obtained by selling x units of A and y units of B.

⇒ Z = 60x + 50y ….(6)

• Hence, the problem is represented mathematically by (2), (3), (4), (5), and (6).
• So, the correct answer is option 1.

Concept

Convert the inequality constraints into equations and find the common points of the bounded region

Calculation

Given LPP

Maximise Z = 40x1 + 50x2

subject to the constraints

x1 + 2x2 ≤ 40,

4x1 + 3x2 ≤ 120,

x1, x2 ≥ 0.

Convert the inequality constraints into equations, we have

x1 + 2x2 = 40,                ..........(1)

4x1 + 3x2 = 120,            ..........(2)

Plotting the Constraints:

1 For x1 + 2x2 = 40

If x₁ = 0 , then 2x₂ = 40 ⇒ x₂ = 20 

If x₂ = 0 , then x₁ = 40 

So, the line passes through points (0, 20) and (40, 0).

For 4x1 + 3x2 = 120:

If x1 = 0 , then 3x2 = 120  ⇒ x2 = 40 

If x2 = 0, then 4x1 = 120  ⇒ x1 = 30 

So, the line passes through points (0, 40) and (30, 0).

Finding Intersection Points:" id="MathJax-Element-110-Frame" role="presentation" style="text-align: center; position: relative;" tabindex="0">

⇒ \( 4(x_1 + 2x_2) = 4 \cdot 40 \Rightarrow 4x_1 + 8x_2 = 160 \)

Now subtract the (2) from this result:

 \((4x_1 + 8x_2) - (4x_1 + 3x_2) = 160 - 120\) 

Simplifying:

⇒ \( 5x_2 = 40 \Rightarrow x_2 = 8 \)

Substitute (x2 = 8) back into the first equation:

⇒ \( x_1 + 2(8) = 40 \Rightarrow x_1 + 16 = 40 \Rightarrow x_1 = 24\) 

Therefore, the intersection point is (24, 8).

Vertices of the Feasible Region:

The vertices of the feasible region are:

(0, 0) (intersection of x1 = 0  and x2 = 0 )

(0, 20) (intersection of x1 = 0 and x1 + 2x2 = 40)

(30, 0) (intersection of x₂ = 0 and 4x1 + 3x2 = 120

(24, 8) (intersection of x1 + 2x2 = 40 and 4x1 + 3x2 = 120

Evaluate Objective Function ( Z ) at Each Vertex:

At (0, 0) ⇒ Z = 40(0) + 50(0) = 0 

At (0, 20) ⇒ Z = 40(0) + 50(20) = 1000 

At (30, 0) ⇒ Z = 40(30) + 50(0) = 1200 

At (24, 8) ⇒ Z = 40(24) + 50(8) = 960 + 400 = 1360 

Therefore, the optimal solution is at (24, 8) with the maximum value (Z = 1360).

∴ Z = 1360 for x1 = 24, x2​ = 8

Explanation:

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6,0), (6, 8), and (0,5).
F = 4x + 6y

Maximum of F – Minimum of F = 72 - 12 = 60 

The Correct option is (1).

Explanation:

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6,0), (6, 8) and (0,5).
F = 4x + 6y

So the Minimum value of F occurs at any point on the line segment joining the points (0, 2) and (3, 0).

The Correct Option is (4).

Explanation:

Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6,0), (6, 8) and (0,5).
F = 4x + 6y

So the Minimum value of F occurs at any point on the line segment joining the points (0, 2) and (3, 0).

The Correct Option is (4).