MATHS M.C.Q (1 Marks)

2. The quantity in column B is greater.

Solution:

Corner points	Corresponding value of Z = 4x + 3y
(0, 0)	0
(0, 40)	120
(20, 40)	200
(60, 20)	300 (Maximum)
(60, 0)	240

Hence, maxmimum value of Z = 300 < 325
So, the quantity in column B is greater.

2. The value of the objective function for a maximization problem will likely be less than that for the simplex solution.

Solution:
Solving an integer programming problem by rounding off answers obtained by solving it as a linear programming problem, we find that the value of the objective function for a maximization problem will likely be less than that for the simplex solution.

3. Corner points of the feasible region

Solution:
Any point in the feasible region that gives the optimal value (maximum or minimum) of the objective function is called an optimal solution.

1. The primal and dual have equal number of variables.

Solution:
The number of variables in dual is equal to the number of constraints in the primal and the number of variables in primal is equal to the number of constraints in the dual.
Therefore, the primal and dual doesnt have equal number of variables.

4. Only firstquadrant.

4. Determine how optimal solution to LPP changes in response to problem inputs.

Solution:
A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.

3. Which are corner points of the feasible region

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. -16

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

Corner points	Corresponding value of z = 3x - 4y
(0, 0)	0
(12, 6)	12 (masimum)
(0, 4)	-16 (miminum)

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

1. Have no feasible solution

Solution:
If there is no point in the feasible set, there is no feasible solution of the linear programming model.
In linear programming, lack of points for a solution set is said to have no feasible solution.

2. Transportation models

Solution:
In linear programming, transportation model are applied to problems related to the study of efficient transportation routes.
For oil companies, how effectively the available resources are transported to different destinations with minimum cost.

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.

1. The problems is to be re - evaluated.

5. Unlimited

Solution:
There is no limit on constraints allowed in linear programming.
so the number of constraints is unlimited.

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. 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.

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.

3. bounded in first quadrant

Solution:
Converting the given inequations into equations, we obtain
y = 6, x + y = 3, x = 0 and y = 0
y = 6 is the line passing through (0, 6) and parallel to the X axis.
The region below the line y = 6 will satisfy the given inequation.
The line x + y = 3 meets the coordinate axis at A(3, 0) and B(0, 3).
Join these points to obtain the line x + y = 3
Clearly, (0, 0) satisfies the inequation x + y ≤ 3.
So, the region in xy-plane that contains the origin represents the solution set of the given equation.
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.
These lines are drawn using a suitable scale.

The shaded region represents the feasible region of the given LPP, which is bounded in the first quadrant.

3. Atleast one of the corner points.

Solution:
Let Z = ax + by be the objective function
When Z has optimum value(maximum or minimum), where the variables
x and y are subject to constraints described by linear inequalities, this optimum value must occur at a corner points of the feasible region.
Thus, the function attains its optimum value at one of the corner points.

2. (0, 8)

4. q = 3p

2. 60 at (4, 2)

2. 600

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.

3. Two

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

1. A maximization function

Solution:
Linear programming problem may be defined as the problem of maximizing or minimizing a linear function subject to linear constraints.
The objective function in a linear program is a maximization function.

3. 37 at (4, 5)

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

4. 36 at (0, 6)

4. None of these

4. None of these

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.

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.

2. (2, 5)

2. 60 at (4, 2)

4. none of these

Solution:
Consider, 2x + 3y = 18

x	y	(x, y)
0	6	(0, 6)
9	0	(9, 0)

Consider, x + y = 10

x	y	(x, y)
0	10	(0, 10)
10	0	(10, 0)

From the graph we conclude that no feasible region exist.

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.