Maths M.C.Q (1 Marks)

2. x₁ = 2, x₂ = 6, Z = 36

Solution:

We need to maximize the function Z = 3x₄ + 5x₂

First, we will convert the given inequations into equations, we obtain the following equations:

3x₁ + 2x₂ = 18, x₁ = 4, x₂ = 6, x₁ = 0 and x₂ = 0

Region represented by 3x₁ + 2x₂ ≤ 18:

The line 3x₁ + 2x₂ = 18 meets the coordinate axes at A(6, 0) and B(0, 9) respectively.

By joining these points we obtain the line 3X1 + 2x2 = 18.

Clearly (0, 0) satisfies the inequation 3x₁ + 2x₂ = 18.

So the region in the plane which contain the origin represents the solution set of the inequation 3x₁ + 2x₂ ≤ 18.

Region represented by x₁ ≤ 4:

The line x₁ = 4 is the line that passes through C(4, 0) and is parallel to the Y axis.

The region to the left of the line x₁ = 4 will satisfy the inequation x₁ ≤ 4.

Region represented by x₂ ≤ 6:

The line x₂ = 6 is the line that passes through D(0, 6) and is parallel to the X axis.

The region below the line x₂ = 6 will satisfy the inequation X₂ ≤ 6.

Region represented by x₁ ≥ 0 and x₂ ≥ 0:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations x₁ ≥ 0 and x₂ ≥ 0.

The feasible region determined by the system of constraints, 3x₁ + 2x₂ ≤ 18, x₁ ≤ 4, x₂ ≤ 6, x₁ ≥ 0 and x₂ ≥ 0 are as follows

Corner points are O(0, 0), D(0, 6), F(2, 6), E(4, 3) and C(4, 0).

The values of the objective function at these points are given in the following table.

We see that the maximum value of the objective function Z is 36 which is at F(2, 6).

3. 37 at (4, 5)

Solution:

Find the maximum value of Z = 3x + 5y referring to the explanation of Q.5.

4. 95 at (2, 3)

3. Which are corner points of the feasible region

4. q = 3p

Solution:

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) + 7(5)

= 3p + 4q = 5q

= q = 3p

3. at any vertex of feasible region

Solution:

In linear programming problem we substitute the coordinates of vertices of feasible region in the objective function and then we obtain the maximum or minimum value.

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

1. Half plane not containing the origin

1. First quadrant

Solution:

 All the positive values of x and y will lie in the first quadrant.

1. It cannot be determined in a graphical solution of an LPP. 

Solution:

There are various methods to solve the linear programming problems namely simplex method, ellipsoid method, graphical method, interior points method, etc.

Therefore a linear programming problem can be solved using the graphical method.

Hence, the feasibility of the linear programming problem can be determined by the graphical method.

4. None of these

1. Feasible solutions

Solution:

In order for a linear programming problem to have a unique solution, the solution must exist at the intersection of two or more constraints.

Then the problem becomes convex and has a single optimum(maximum or minimum) solution.

Therefore the convex set of equations is included in the feasible region.

1. Alternate courses of action to choose from.

Solution:

According to Robbins, the resources(capital, land, labour, materials, ...) are always limited.

Every resource have multiple uses.

The problem before any organisation or manager is to choose the best alternatives which can maximize the profit or minimize the cost of production.

Linear programming is the method which is used to select the best possible alternatives from the all alternatives.

According to William M. Fox, "Linear programming is a planning technique that permits some objective function to be maximized or minimized within the framework of given situational restrictions"

Therefore, the linear programming is the process of selecting best courses of action to choose from various alternatives.

1. The problem is infeasible

Solution:

Any linear programming problem must have the following properties:-1.

The relationship between variables and constraints must be linear2.

The constraints must be non - negative.3.. objective function must be linear.

Non - negativity conditions are used because the variables cannot take negative values.

i.e., it is not possible to have negative resources (land, capital, labour cannot be negative).

Because of the non - negativity condition, the feasible region exists only in I quadrant.

2. 60 at (4, 2)

4. none of these

Solution:

Consider, 2x + 3y = 18

Consider, x + y = 10

From the graph we conclude that no feasible region exist.

3. Destinations

Solution:

In linear programming, transportation modeltransportation model are applied to problems related to the study of efficient transportation routes.

i.e., how effectively the available resources are transported to different destinations with minimum cost.

Therefore, the points of demand is classified as destinations.

2. a function to be optimized

Solution:

The objective function of a linear programming problem is either to be maximized or minimized i.e. objective function is to be optimized.

4. none of these

Solution:

We need to maximize the function Z = 4x + 3y

Converting the given inequations into equations, we obtain

3x + 2y = 160, 5x + 2y = 200, x + 2y = 80, x = 0 and y = 0

Region represented by 3x + 2y ≥ 160:

The line 3x + 2y = 160 meets the coordinate axes at A1603,0 and B(0, 80) respectively.

By joining these points we obtain the line 3x + 2y = 160.

Clearly (0, 0) does not satisfies the inequation 3x + 2y ≥ 160.

So, the region in xy plane which does not contain the origin represents the solution set of the inequation 3x + 2y ≥ 160.

Region represented by 5x +2y ≥ 200:

The line 5x + 2y = 200 meets the coordinate axes at C(40, 0) and D(0, 100) respectively.

By joining these points we obtain the line 5x + 2y = 200.

Clearly (0, 0) does not satisfies the inequation 5x +2y ≥ 200.

So, the region which does not contain the origin represents the solution set of the inequation 5x +2y ≥ 200.

Region represented by x +2y ≥ 80:

The line x + 2y = 80 meets the coordinate axes at E(80, 0) and F(0, 40) respectively.

By joining these points we obtain the line x + 2y = 80.

Clearly (0, 0) does not satisfies the inequation x + 2y ≥ 80.

So, the region which does not contain the origin represents the solution set of the inequation x + 2y ≥ 80.

Region represented by x ≥ 0 and y ≥ 0:

Since, every point in the first quadrant satisfies these inequations.

So, the first quadrant is the region represented by the inequations x ≥ 0, and y ≥ 0.

The feasible region determined by the system of constraints 3x + 2y ≥ 160,5x+2y ≥ 200, x +2y ≥ 80, x ≥ 0, and y ≥ 0 are as follows.

Here, we see that the feasible region is unbounded.

Therefore,maximum value is infinity.

3. Two

Solution:

Graphical method can be used only when the decision variables is two.

2. (1.5, 4)

4. Has been formulated improperly.

Solution:

A linear programming problem is said to have unbounded solution if it has infinite number of solutions.

I.e., the problem has been formulated improperly.

2. 600

1. (3.5, 0)

3. Convex Polygon

Solution:

A bounded feasible region will have both a maximum value and a minimum value for the objective function. It is bounded if it can be enclosed in any shape.

A convex polygon is a simple not self-intersecting closed shape in which no line segment between two points on the boundary ever goes outside the polygon.

Hence, the answer is convex polygon.

2. -16

Solution:

the feasible region as show in the figure, has objective function F= 3x - 4y

We have minimum value of F is -16at (0, 4).

1. y = 4 + 6x

Solution:

First kmkm fare = Rs.10 Subsequent distance fare = Rs 6/ km

Then fare x km of distance y = (x - 1) × 6 + 10y = 6x - 6 + 10y = 6x + 4

2. (0, 8)

4. All of the above

4. None of the above

Solution:

If an iso profit line which is yielding the optimal solution coincide with a constant line; then

→ the solution will b bounded, i.e there will be a definite bounded area where the solution would be optional.

→ Since the area is bounded,the solution is feasible

→ And the constant which coincides is not a redundant

Hence None of above is the answer.

2. Corner points

Solution:

The fundamental theorem of programming (i.e., Convex Polygon Theorem) states that the optimum value(maximum or minimum) of a linear programming problem over a convex region occur at the corner points.

1. Feasible

Solution:

Our experts are building a solution for this.

1. Both A and R are true and R is the correct explanation of A

2. 60 at (4, 2)

1. {x : |x| = 5}

Solution:

|x| = 5 is not a convex set as any two points from negative and positive x-axis if are joined will not lie in set.

4. Infinite

1. The problems is to be re - evaluated.