Case study (4 Marks)

Question 14 Marks

Let R be the feasible region (convex polygon) for a linear programming problem and let Z = ax + by be the objective function. When Z has an optimal value (maximum or minimum), where the variables x and y are subject to constraints described by linear inequalities, this optimal value must occur at a corner point (vertex) of the feasible region.

Based on the above information, answer the following questions.

Objective function of a L.P.P. is:

A constant.
A function to be optimised.
A relation between the variables.
None of these.

Which of the following statement is correct?

Every LPP has at least one optimal solution.
Every LPP has a unique optimal solution.
If an LPP has two optimal solutions, then it has infinitely many solutions.
None of these.

In solving the LPP: "minimize f = 6x + 10y subject to constraints $\text{x}\geq6,\text{ y}\geq2,\text{ 2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0"$ redundant constraints are:

$\text{x}\geq6,\text{ y}\geq2$
$\text{2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0$
$\text{x}\geq6$
None of these

The feasible region for a LPP is shown shaded in the figure. Let Z = 3x - 4y be the objective function. Minimum of Z occurs at:

(0, 0)
(0, 8)
(5, 0)
(4, 10)

The feasible region for a LPP is shown shaded in the figure. Let F = 3x - 4y be the objective function. Maximum value of F is:

0
8
12
-18

Answer

(b) A function to be optimised.

Solution:

Objective function is a linear function (involve variable) whose maximum or minimum value is to be found.

(c) If an LPP has two optimal solutions, then it has infinitely many solutions.

Solution:

If optimal solution is obtained at two distinct points A and B ( corners of the feasible region), then optimal solution is obtained at every point of segment [AB].

(b) $\text{2x}+\text{y}\geq10,\text{ x}\geq0,\text{ y}\geq0$

Solution:

When $\text{x}\geq6$ and $\text{y}\geq2,$ then

$\text{2x}+\text{y}\geq2\times6+2,\text{i.e.,}\text{ 2x}+\text{y}\geq14$

Hence, $\text{x}\geq0,\text{ y}\geq0$ and $2\text{x}+\text{y}\geq10$ are automatically satisfied by every point of the region

$\{(\text{x, y}):\text{x}\geq6\}\cap\{(\text{x, y}):\text{y}\geq2\}$

(b) (0, 8)

Solution:

Construct the following table of values of the objective function:

Corner Point	Value of Z = 3x - 4y
(0, 0)	3 × 0 - 4 × 0 = 0
(5, 0)	3 × 5 - 4 × 0 = 15
(6, 5)	3 × 6 - 4 × 5 = -2
(6, 8)	3 × 6 - 4 × 8 = -14
(4, 10)	3 × 4 - 4 × 10 = -28
(0, 8)	$3\times0 - 4\times8 = -32\leftarrow\text{Minimum}$

Minimum of Z = -32 at (0, 8).

(a) 0

Solution:

Construct the following table of values of the objective function F:

Corner Point	Value of F = 3x - 4y
(0, 0)	$3\times0 - 4\times0 = 0\leftarrow\text{Minimum}$
( 6, 12)	3 × 6 - 4 × 12 = -30
(6, 16)	3 × 6 - 4 × 16 = -46
(0, 4)	3 × 0 - 4 × 4 = -16

Hence, maximum of F = 0.

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Question 24 Marks

Deepa rides her car at 25 km/ hr. She has to spend ₹ 2 per km on diesel and if she rides it at a faster speed of 40 km/ hr, the diesel cost increases to ₹ 5 per km. She has ₹ 100 to spend on diesel. Let she travels x kms with speed 25 km/ hr and y kms with speed 40 km/ hr. The feasible region for the LPP is shown below:

Based on the above information, answer the following questions.

What is the point of intersection of line l₁ and l₂?

$\Big(\frac{40}{3},\frac{50}{3}\Big)$
$\Big(\frac{50}{3},\frac{40}{3}\Big)$
$\Big(\frac{-50}{3},\frac{40}{3}\Big)$
$\Big(\frac{-50}{3},\frac{-40}{3}\Big)$

The comer points of the feasible region shown in above graph are:

$(0,25),(20,0),\Big(\frac{40}{3},\frac{50}{3}\Big)$
$(0, 0), (25, 0), (0, 20) $
$(0,0),\Big(\frac{40}{3},\frac{50}{3}\Big),(0,20)$
$(0,0),(25,0),\Big(\frac{50}{3},\frac{40}{3}\Big),(0,20)$

If Z = x + y be the objective function and max Z = 30. The maximum value occurs at point:

$\Big(\frac{50}{3},\frac{40}{3}\Big)$
(0, 0)
(25, 0)
(0, 20)

If Z = 6x - 9y be the objective function, then maximum value of Z is:

-20
150
180
20

If Z = 6x + 3y be the objective function, then what is the minimum value of Z?

120
130
0
150

Answer

(b) $\Big(\frac{50}{3},\frac{40}{3}\Big)$

Solution:

Let B (x, y) be the point of intersection of the given lines 2x + Sy = 100

And $\frac{\text{x}}{25}+\frac{\text{y}}{40}=1$

⇒ 8x + 5y = 20

Solving (i) and (ii), we get

$\text{x}=\frac{50}{3},\text{y}=\frac{40}{3}$

$\therefore$ The point of intersection B(x, y) $$

$=\Big(\frac{50}{3},\frac{40}{3}\Big).$

(d) $(0,0),(25,0),\Big(\frac{50}{3},\frac{40}{3}\Big),(0,20)$

Solution:

The corner points of the feasible region shown in the given graph are:

$(0,0),\text{A}(25,0),\text{B}\Big(\frac{50}{3},\frac{40}{3}\Big),\text{C}(0,20)$

(a) $\Big(\frac{50}{3},\frac{40}{3}\Big)$

Solution:

Here Z = x + y

Corner Points	Value of Z = x + y
(0, 0)	0
(25, 0)	$$25
$\Big(\frac{50}{3},\frac{40}{3}\Big)$	$30\leftarrow\text{Maximum} $
(0, 20)	20

Thus, max Z = 30 occurs at point $\Big(\frac{50}{3},\frac{40}{3}\Big).$

(b) 150

Solution:

Corner Points	Value of Z = 6x - 9y
(0, 0)	0
(25, 0)	$150\leftarrow\text{Maximum} $
$\Big(\frac{50}{3},\frac{40}{3}\Big)$	-20
(0, 20)	-180

(c)

Solution:

Corner Points	Value of Z = 6x +3y
(0, 0)	$0\leftarrow\text{Maximum} $
(25, 0)	150
$\Big(\frac{50}{3},\frac{40}{3}\Big)$	140
(0, 20)	60

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Question 34 Marks

Suppose a dealer in rural area wishes to purpose a number of sewing machines. He has only ₹ 5760 to invest and has space for at most 20 items for storage. An electronic sewing machine costs him ₹ 360 and a manually operated sewing machine ₹ 240. He can sell an electronic sewing machine at a profit of ₹ 22 and a manually operated sewing machine at a profit of ₹ 18.

Based on the above information, answer the following questions.

Let x and y denotes the number of electronic sewing machines and manually operated sewing machines purchased by the dealer. If it is assume that the dealer purchased atleast one of the the given machines, then:

$\text{x}+\text{y}\geq0$
$\text{x}+\text{y}<0$
$\text{x}+\text{y}>0$
$\text{x}+\text{y}\leq0$

Let the constraints in the given problem is represented by the following inequalities.

$\text{x}+\text{y}\leq20$

$360\text{x}+240\text{y}\leq5760$

$\text{x},\text{y}\geq0$

Then which of the following point lie in its feasible region.

(0, 24)
(8, 12)
(20, 2)
None of these

If the objective function of the given problem is maximise z = 22x + 18y, then its optimal value occur at:

(0, 0)
(16, 0)
(8, 12)
(0, 20)

Suppose the following shaded region APDO, represent the feasible region corresponding to mathematical formulation of given problem.

Then which of the following represent the coordinates of one of its corner points

(0, 24)
(12, 8)
(8, 12)
(6, 14)

If an LPP admits optimal solution at two consecutive vertices of a feasible region, then:

The required optimal solution is at the midpoint of the tine joining two points.
The optimal solution occurs at every point on the tine joining these two points.
The LPP under consideration is not solvable.
The LPP under consideration must be reconstructed.

Answer

(c) $\text{x}+\text{y}>0$
(b) (8, 12)

Solution:

Since (8, 12) satisfy all the inequalities, therefore (8, 12) is the point in its feasible region.

(c) (8, 12)

Solution:

At (0, 0), z = 0

At (16, 0), z = 352

At (8, 12), z = 392

At (0, 20), z = 360

It can be observed that max z occur at (8, 12). Thus, z will attain its optimal value at (8, 12).

(c) (8, 12)

Solution:

We have, x + y = 20 (i)

And 3x + 2y = 48 (ii)

On solving (i) and (ii), we get

x = 8, y = 12.

Thus, the coordinates of Pare (8, 12) and hence (8, 12) is one of its corner points.

(b) The optimal solution occurs at every point on the tine joining these two points.

Solution:

The optimal solution occurs at every point on the line joining these two points.

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Question 44 Marks

Linear programming is a method for finding the optimal values (maximum or minimum) of quantities subject to the constraints when relationship is expressed as linear equations or inequations.

Based on the above information, answer the following questions.

The optimal value of the objective function is attained at the points:

On X-axis.
On Y-axis.
Which are comer points of the feasible region.
None of these.

The graph of the inequality 3x + 4y < 12 is:

Half plane that contains the origin.
Half plane that neither contains the origin nor the points of the line 3x + 4y = 12.
Whole XOY-plane excluding the points on line 3x + 4y = 12.
None of these.

The feasible region for an LPP is shown in the figure. Let Z = 2x + 5y be the objective function. Maximum of Z occurs at:

(7, 0)
(6, 3)
(0, 6)
(4, 5)

The corner points of the feasible region determined by the system of linear constraints are (0, 10), (5, 5), (15, 15), (0, 20). Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both the points ( 15, 15) and (0, 20) is:

p = q
p = 2q
q = 2p
q = 3p

The comer points of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y.

Compare the quantity in Column A and Column B

Column A	Column B
Maximum of Z	325

The quantity in column A is greater.
The quantity in column Bis greater.
The two quantities are equal.
The relationship cannot be determined on the basis of the information supplied.

Answer

(c) Which are comer points of the feasible region.

Solution:

When we solve an L.P.P. graphically, the optimal (or optimum) value of the objective function is attained at comer points of the feasible region.

(d) None of these.

Solution:

From the graph of 3x + 4y < 12, it is clear that it contains the origin but not the points on the line 3x + 4y = 12.

(d) (4, 5)

Solution:

Maximum of objective function occurs at corner points.

Corner Points	Value of Z = 2x + 5y
(0, 0)	0
(7, 0)	14
(6, 3)	27
(4, 5)	$33\leftarrow\text{Maximum}$
(0, 6)	30

(d) q = 3p

Solution:

Value of Z = px + qy at ( 15, 15)= 15p + 15q and that at (0, 20) = 20q. According to given condition, we have 15p + 15q = 20q ⇒ 15p = Sq ⇒ q = 3p.

(b) The quantity in column Bis greater.

Solution:

Construct the following table of values of the objective function:

Corner Points	Value of Z = 4x + 3y
(0, 0)	4 × 0 + 3 × 0 = 0
(0, 40)	4 × 0 + 3 × 40 = 120
(20, 40)	4 × 20 + 3 × 40 = 200
(60, 20)	$4\times60+3\times20=300\leftarrow\text{Maximum}$
(60, 0)	4 × 60 + 3 × 0 = 240

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Question 54 Marks

Corner points of the feasible region for an LPP are (0, 3), (5, 0), (6, 8), (0, 8). Let Z = 4x - 6y be the objective function.

Based on the above information, answer the following questions.

The minimum value of Z occurs at:

(6, 8)
(5, 0)
(0, 3)
(0, 8)

Maximum value of Z occurs at:

(5, 0)
(0, 8)
(0, 3)
(6, 8)

Maximum of Z - Minimum of Z =

58
68
78
88

The corner points of the feasible region determined by the system of linear inequalities are:

(0, 0), (-3, 0), (3, 2). (2, 3)
(3, 0), (3, 2), (2, 3), (0, -3)
(0, 0), (3, 0), (3, 2), (2, 3), (0, 3)
None of these

The feasible solution of LPP belongs to:

First and second quadrant.
First and third quadrant.
Only second quadrant.
Only first quadrant.

Answer

Construct the following table of values of objective function:

Corner Points	Value of Z = 4x - 6y
(0, 3)	4 × 0 - 6 × 3 = -18
(5, 0)	4 × 5 - 6 × 0 = 20
(6, 8)	4 × 6 - 6 × 8 = -24
(0, 8)	4 × 0 - 6 × 8 = -48

(d) (0, 8)

Solution:

Minimum value of Z is -48 which occurs at (0, 8).

(a) (5, 0)

Solution:

Maximum value of Z is 20, which occurs at (5, 0).

(b) 68

Solution:

Maximum of Z - Minimum of Z = 20 - (-48) = 20 + 48 = 68

(c) (0, 0), (3, 0), (3, 2), (2, 3), (0, 3)

Solution:

The comer points of the feasible region are O(0, 0), A(3, 0), B(3, 2), C(2, 3), D(0, 3).

(d) Only first quadrant.

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Maths Case study (4 Marks)

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