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Let x units of fodder 1 and y units of fodder 2 be prescribed.

The cost of fodder 1 is Rs 3 per unit and cost of fodder 2 is Rs 2 per unit.

∴ total cost is z = 3x + 2y

This is the linear function which is to be minimized. Hence it is the objective function. 

The constraints are as per the following table :

From table fodder contains (2x + y) units of nutrients A, (2x + 3y) units of nutrients B and (x + y) units of nutrients C. The minimum requirements of these nutrients are 14 units, 22 units and 1 unit respectively.

Therefore, the constraints are

2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1 

Since, number of units (i.e. x and y) cannot be negative, we have, x ≥ 0, y ≥ 0. 

Hence, the given LPP can be formulated as Minimize z = 3x + 2y, subject to 2x + y ≥ 14, 2x + 3y ≥ 22, x + y ≥ 1, x ≥ 0, y ≥ 0.

Let the units of food F, be x and food F, be y minimize 

z = 50x + 75y 

subject to the restrictions 

2x + 3y ≥ 6 

3x + 4y ≥ 8 x,y ≥ 0

Correct answer is (D) (40, 15)

Let the number of units of product A be x & product B be y

Maximize z = 2x + 3y 

subject to the constraints [6 hrs + 40 minutes 360 minutes + 40 minutes = 400 minutes] 

x + y ≤ 400 

2x + y ≤ 600 

x, y ≥ 0

We have to maximize the profit by calculating the number of units needed to produce each product.

For Product A,

Cost Price per unit = Labour Cost + Raw material cost per unit

Cost Price per unit = 16 + 4 = Rs. 20

Selling Price per unit = Rs. 25

Profit per unit = S.P – C.P = 25 – 20 = Rs. 5

For Product B,

Cost Price per unit = Labour Cost + Raw material cost per unit

Cost Price per unit = 20 + 4 = Rs. 24

Selling Price per unit = Rs. 30

Profit per unit = S.P – C.P = 30 – 24 = Rs. 6

Let number of units produced of Product A be x and number of units produced of Product B be y.

Hence, Total Profit = 5 x + 6 y

To Maximize : z = 5 x + 6 y

For Department 1,

3 x + 2 y ≤ 130

For Department 2,

4 x + 6 y ≤ 260

Hence,

Z = 5 x + 6 y

3 x + 2 y ≤ 130

4 x + 6 y ≤ 260

x, y ≥ 0 [Since production cannot be less than zero]

The maximum value of Z is unique.
It is given that the maximum value of Z occurs at two points, (3, 4) and (0, 5).
∴ Value of Z at (3, 4) = Value of Z at (0, 5)

⇒ p(3) + q(4) = p(0) + q(5)
⇒ 3p + 4q = 5q
⇒ q = 3p
Hence, the correct answer is D.

A comer point of a feasible region is a point in the region which is the intersection of two boundary lines.

In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

Answer is (D)

At (0, 0) Z = 0 

At (5, 0) Z = 5p 

At (3, 4) Z = 3p +4q 

At (0, 5) Z = 5q 

∴ maximum of Z 

3p + 4q = 5q 

∴ 3p = q

A feasible region of a system of linear inequalities is said to be bounded, if it can be enclosed within a circle.

In a LPP, the linear inequalities or restrictions on the variables are called linear constraints.

A feasible region of a system of linear inequalities is said to be bounded if it can be enclosed within a circle.

The feasible region for an LPP is always a convex polygon.

Answer is True

In a LPP, the objective function is always linear.

In a LPP if the objective function Z = ax + by has the same maximum value on two corner points of the feasible region, then every point on the line segment joining these two points give the same maximum value

True.

If the feasible region is unbounded then we may or may not have a maximum or minimum of objective function, but if we have a maximum or a minimum value then it must be at one of the corner points only.

If the feasible region for a LPP is unbounded, then the optimal value of the objective function Z = ax + by may or may not exist.

In a LPP, objective function is always linear.

If the feasible region for a LPP is unbounded, then the optimal value of the objective function Z = ax + by may or may not exist.

Correct answer is B.

Given,

To define the objective of a Linear programming Problem.

As per the definition of the Linear Programming Problem,

A Linear programming problem is a linear function (also known an objective function) subjected to certain constraints for which we need to find an optimal solution (i.e. either a maximum/minimum value) depending on the requirement of the problem.

From the above definition, we can clearly say that, Linear programming problem’s objective is to either maximize/ minimize a given objective function, which means to optimize a function to get an optimum solution.

Correct answer is C.

Given,

The statements:

• Every LPP admits an optimal solution

This need not be true as all LPPs need not have optimal solutions and such LPPs are called unbound.

• A LPP admits unique optimal solution

Every LLP need not have unique optimal solutions as if there are two optimal solutions to an LLP there will be infinite number or optimal solutions to the LLP problem.

• If a LPP admits two optimal solutions it has an infinite number of optimal solutions

As mentioned in the above point, if there are two optimal solutions to an LLP there will be infinite number or optimal solutions to the LLP problem.

• The set of all feasible solutions of a LPP is not a convex set.

As per a theorem of Convex Sets,

If {X₁, X₂} ∈ C (a convex set of optimal solutions), then

X = λX₁ + (1 − λ) X₂ where 0 ≤ λ ≤ 1, is also contained in C (the optimal solution set). This makes all the feasible solutions of a LPP also a convex set.

Correct answer is A.

Given,

The constraints of a linear programming problem are changed.

Now, as per the definition of the Linear Programming Problem,

A Linear programming problem is a linear function (also known an objective function) subjected to certain constraints for which we need to find an optimal solution (i.e. either a maximum/minimum value) depending on the requirement of the problem.

Here, the LPP is solved using the constraints, so, if the constraints are changed, the problem is to has to be re-calculated with the new constraints provided.

Correct answer is C.

Given that,

The objective function will be maximum at ______

As per the steps while solving a Linear Programming Problem, we first determine the feasible region using the constraints and then, consider the vertices of the obtained convex polygon in the objective function to see the vertex at which the objective function obtains a maximum value. Then that vertex is determines the maximum value of the objective function.

Hence, the maximum value of an objective function under linear constraints is at any vertex of the feasible region.

Correct answer is (C) at a vertex of feasible region

Correct answer is (B) (0, 0), (7/2, 0), (3, 1), (0, 4)

(i) From the figure the feasible region is OABC. Then the comer points are; A is (5, 0), B is (3, 4), C is (0, 5) and O (0, 0) 

(ii) The constraints are 2x + y < 10, x + 3y < 15, x < 0, y < 0 

(iii) Given; Z = px + qy

Since maximum at A and B we have; 

⇒ 3p + 4q = 5p ⇒ 2p = 4q ⇒ p = 2q 

(iv) When q = 1, then p ⇒ 2q ⇒ p = 2 

Objective function is; Z = 2x + y 

(v) We have; Z px + qy at B Z has maximum 

⇒ Z = 2(3) + 4 

= 10

Let x be the running time for machine A and y be the running time for machine B. Since machines cannot work more than 12 hours x + y < 12 

Since maximum production of two machines is 800 units. 60x + 80y < 800 

Maximum cost of production is 25000, 

2000x + 2500y < 25000 0 < x < 10, 0 < y < 7

(i) Let x be the number of packages of nuts

produced and y be the number of packages of bolts produced. Then; Maximise profit is; Z = 17. 5x + 7y

(ii) Time constraint for Machine A; x + 3y < 12 Time constraint for Machine B; 3x + y < 12 Therefore; Maximise; Z = 17.5x + 7y, x + 3y < 12, 3x + y < 12, x < 0, y < 0

(iii) The shaded region OABC is the visible region. Here the region is bounded. The corner points are 0(0,0), A (4, 0) B(3, 3), C(0, 4).

Given; Z = 17.5x + 7y

Since maximum value of Z occurs at B, the solution is 

Z = 17.5(3) + 7(3) 

= 73.5.

Answer is False

Answer is False

Answer is False

The minimum value can also occur at more than one corner points of the feasible region.

Answer is Feasible.

Answer is True

Let the number of units of nuts x & bolts be y it takes 1 hour on machine A for a nut and 3 hours on machine 13 for a nut maximum time available for machine A is 12 hours. 

∴ x + 3y ≤ 12 

3x + y ≤ 12 

Negative units cannot be produced: x, y ≥ 0 our objective is to maximum profit the LPP is Maximize 

= 2.5 x + y subject to 

x + 3y ≤ 12 

3x + y ≤ 12 

and x ≥ 0, y ≥ 0

Correct answer is (B) (2, 0)

Correct answer is (A) (2, 2)

Correct answer is (C) (0, 0)

Correct answer is (C) (3, 4)

(D⁴ + 2D³ – 3D² – 4D + 4)y = 0

Its auxilary equation is

m⁴ + 2m³ – 3m² – 4m + 4 = 0

⇒ (m – 1)(m³ + 3m² – 4) = 0

⇒ (m – 1)²(m² + 4m + 4) = 0

⇒ (m – 1)²(m + 2)² = 0

⇒ m = 1, 1, –2, –2.

C.F. is y = (C₁ + C₂x)e^x + (C₃ + C₄x)e^-2x

Hence, solution of given differential equation is y = (C₁ + C₂x)e^x + (C₃ + c₄x)e^-2x.

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Let’s assume x and y to be the number of sweaters of type A and type B respectively.

From the question, the following constraints are:

360x + 120y ≤ 72000 ⇒ 3x + y ≤ 600 … (i)

x + y ≤ 300 … (ii)

x + 100 ≥ y ⇒ y ≤ x + 100 … (iii)

Profit: Z = 200x + 120y

Therefore, the required LPP to maximize the profit is

Maximize Z = 200x + 120y subject to constrains

3x + y ≤ 600, x + y ≤ 300, y ≤ x + 100, x ≥ 0, y ≥ 0.

Let the company manufactures x number of type A sweaters and y number of type B.

The company spend at most Rs 72000 a day.

∴ 360x + 120y ≤ 72000

=> 3x+y≤ 600 …(i)

Also, company can make at most 300 sweaters.

∴ x+y≤ 300 …(ii)

Also, the number of sweaters of type B cannot exceed the number of sweaters of type A by more than 100 i.e., y-x≤ 100

The company makes a profit of Rs 200 for each sweater of type A and Rs 120 for every sweater of type B

So, the objective function for maximum profit is Z = 200x + 120y subject to constraints. 3x+y≤ 600

x+y ≤ 300

x-y ≥ -100

x ≥ 0, y ≥ 0

Let the number of large vans and small vans be x and y respectively.

here transportation cost Z be objective be function,

Z = 400x + 200y which is to be minimized under constraints

200 x + 80y ≥ 1200 ⇒ 5x + 2y ≥ 30

400 x + 200y ≤ 3000 ⇒ 2x + y ≤ 15

x ≤ y, x ≥ 0, y ≥ 0

B. (0, 8)

Value of Z = 3x – 4y, at corner points are –

At (0, 0) = 0

At (0, 8) = -32

At (4, 10) = -28

At (6, 8) = -14

At (6, 5) = -2

At (5, 0) = 15

So, clearly minimum value is at (0, 8)

Value of Z = 3x – 4y, at corner points are –

At (0, 0) = 0

At (0, 8) = -32

At (4, 10) = -28

At (6, 8) = -14

At (6, 5) = -2

At (5, 0) = 15

So, clearly maximum value is at (5, 0)

Value of Z = 3x – 4y, at corner points are –

At (0, 0) = 0

At (0, 8) = -32

At (4, 10) = -28

At (6, 8) = -14

At (6, 5) = -2

At (5, 0) = 15

So, clearly maximum value is at (5, 0)

Maximum value = 15,

Minimum value = -32

So, maximum + minimum = 15 -32

= -17

F = 3x – 4y

Corner points are (0, 0), (0, 4), (12, 6)

Value of F at corner points –

At (0, 0), F = 0

At (0, 4), F = -16

At (12, 6), F = 12

Clearly, maximum value of F = 12.

F = 3x – 4y

Corner points are (0, 0), (0, 4), (12, 6)

Value of F at corner points –

At (0, 0), F = 0

At (0, 4), F = -16

At (12, 6), F = 12

Clearly, minimum value of F = – 16.

The correct answer is (B).