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Linear Inequalities and Linear Programming

88622 Consider a Linear Programming Problem: Minimize $Z = 5 x + 3 y$ , Subject to : $3 x + y \geq 10$ , $2 x + 2 y \geq 14$ and $x + 2 y \geq 9$ . Which one of the following points lies outside the feasible region?

1

(1, 9)

2

(4, 2)

3

(6, 2)

4

(12, 2)

Explanation:

(B) : Given,
Minimize, $Z = 5 x + 3 y$
Subject to
$3 x + y \geq 10$
$2 x + 2 y \geq 14$
And $x + 2 y \geq 9$
On plotting the constraints, we get-
$A (9, 0), B (5, 2), C (\frac{3}{2}, \frac{11}{2})$ and $D (0, 10)$ as corner
point of feasible region which is unbounded.
Now, $(1, 9), (6, 2)$ and $(12, 2)$ lies inside the feasible regions and point $(4, 2)$ lies outside the feasible region.

J&K CET-2019

Linear Inequalities and Linear Programming

88623 The corner points of the feasible region determined by the system of linear constraints are $(0, 10), (5, 5), (25, 20)$ and $(0, 30)$ . Let $Z =$ $p x + q y$ , where $p, q > 0$ , Condition on $p$ and $q$ so that the maximum of $Z$ occurs at both the points $(25, 20)$ and $(0, 30)$ is

1

5 p = 2 q

2

2 p = 5 q

3

p = 2 q

4

q = 3 p

Explanation:

(A) : Given,
Let, $z = px + qy$ , where $p, q > 0$
And given maximum of $Z$ occurs at both the points (25, $20)$ and $(0, 30)$ .
Then
${Max}_{(25, 20)} = {MaxZ}_{(0, 30)}$
$25 p + 20 q = 0 + 30 q$
$25 p = 10 q$
Therefore, condition on $p$ and $q$
$5 p = 2 q$

GUJCET-2017

Linear Inequalities and Linear Programming

88624 The coordinates of the corner points of the bounded feasible region are $(0, 10), (5, 5), (15$ , $15)$ and $(0, 20)$ . The maximum of the objective function $Z = 10 x + 20 y$ is:

1 600

2 550

3 400

4 450

Explanation:

(D) :
| Corner points | $Z = 10 x + 20 y$ |
| :--- | :--- |
| $(0, 10)$ | $0 + 200 = 200$ |
| $(5, 5)$ | $10 \times 5 + 20 \times 5 = 150$ |
| $(15, 15)$ | $10 \times 15 + 20 \times 15 =$ |
| | 450 |
| $(0, 20)$ | $0 + 400 = 400$ |
Maximum of the objective function, $x = 450$

GUJCET-2023

Linear Inequalities and Linear Programming

88613 If the number of available constraints is 3 and the number of parameters to be optimized is 4 , then

1 The objective function can be optimized

2 The constraint are short in number

3 The solution is problem oriented

4 None of these

BITSAT-2019

Linear Inequalities and Linear Programming

88622 Consider a Linear Programming Problem: Minimize $Z = 5 x + 3 y$ , Subject to : $3 x + y \geq 10$ , $2 x + 2 y \geq 14$ and $x + 2 y \geq 9$ . Which one of the following points lies outside the feasible region?

1

(1, 9)

2

(4, 2)

3

(6, 2)

4

(12, 2)

Explanation:

(B) : Given,
Minimize, $Z = 5 x + 3 y$
Subject to
$3 x + y \geq 10$
$2 x + 2 y \geq 14$
And $x + 2 y \geq 9$
On plotting the constraints, we get-
$A (9, 0), B (5, 2), C (\frac{3}{2}, \frac{11}{2})$ and $D (0, 10)$ as corner
point of feasible region which is unbounded.
Now, $(1, 9), (6, 2)$ and $(12, 2)$ lies inside the feasible regions and point $(4, 2)$ lies outside the feasible region.

J&K CET-2019

Linear Inequalities and Linear Programming

88623 The corner points of the feasible region determined by the system of linear constraints are $(0, 10), (5, 5), (25, 20)$ and $(0, 30)$ . Let $Z =$ $p x + q y$ , where $p, q > 0$ , Condition on $p$ and $q$ so that the maximum of $Z$ occurs at both the points $(25, 20)$ and $(0, 30)$ is

1

5 p = 2 q

2

2 p = 5 q

3

p = 2 q

4

q = 3 p

Explanation:

(A) : Given,
Let, $z = px + qy$ , where $p, q > 0$
And given maximum of $Z$ occurs at both the points (25, $20)$ and $(0, 30)$ .
Then
${Max}_{(25, 20)} = {MaxZ}_{(0, 30)}$
$25 p + 20 q = 0 + 30 q$
$25 p = 10 q$
Therefore, condition on $p$ and $q$
$5 p = 2 q$

GUJCET-2017

Linear Inequalities and Linear Programming

88624 The coordinates of the corner points of the bounded feasible region are $(0, 10), (5, 5), (15$ , $15)$ and $(0, 20)$ . The maximum of the objective function $Z = 10 x + 20 y$ is:

1 600

2 550

3 400

4 450

Explanation:

(D) :
| Corner points | $Z = 10 x + 20 y$ |
| :--- | :--- |
| $(0, 10)$ | $0 + 200 = 200$ |
| $(5, 5)$ | $10 \times 5 + 20 \times 5 = 150$ |
| $(15, 15)$ | $10 \times 15 + 20 \times 15 =$ |
| | 450 |
| $(0, 20)$ | $0 + 400 = 400$ |
Maximum of the objective function, $x = 450$

GUJCET-2023

Linear Inequalities and Linear Programming

88613 If the number of available constraints is 3 and the number of parameters to be optimized is 4 , then

1 The objective function can be optimized

2 The constraint are short in number

3 The solution is problem oriented

4 None of these

BITSAT-2019

Linear Inequalities and Linear Programming

88622 Consider a Linear Programming Problem: Minimize $Z = 5 x + 3 y$ , Subject to : $3 x + y \geq 10$ , $2 x + 2 y \geq 14$ and $x + 2 y \geq 9$ . Which one of the following points lies outside the feasible region?

1

(1, 9)

2

(4, 2)

3

(6, 2)

4

(12, 2)

Explanation:

(B) : Given,
Minimize, $Z = 5 x + 3 y$
Subject to
$3 x + y \geq 10$
$2 x + 2 y \geq 14$
And $x + 2 y \geq 9$
On plotting the constraints, we get-
$A (9, 0), B (5, 2), C (\frac{3}{2}, \frac{11}{2})$ and $D (0, 10)$ as corner
point of feasible region which is unbounded.
Now, $(1, 9), (6, 2)$ and $(12, 2)$ lies inside the feasible regions and point $(4, 2)$ lies outside the feasible region.

J&K CET-2019

Linear Inequalities and Linear Programming

88623 The corner points of the feasible region determined by the system of linear constraints are $(0, 10), (5, 5), (25, 20)$ and $(0, 30)$ . Let $Z =$ $p x + q y$ , where $p, q > 0$ , Condition on $p$ and $q$ so that the maximum of $Z$ occurs at both the points $(25, 20)$ and $(0, 30)$ is

1

5 p = 2 q

2

2 p = 5 q

3

p = 2 q

4

q = 3 p

Explanation:

(A) : Given,
Let, $z = px + qy$ , where $p, q > 0$
And given maximum of $Z$ occurs at both the points (25, $20)$ and $(0, 30)$ .
Then
${Max}_{(25, 20)} = {MaxZ}_{(0, 30)}$
$25 p + 20 q = 0 + 30 q$
$25 p = 10 q$
Therefore, condition on $p$ and $q$
$5 p = 2 q$

GUJCET-2017

Linear Inequalities and Linear Programming

88624 The coordinates of the corner points of the bounded feasible region are $(0, 10), (5, 5), (15$ , $15)$ and $(0, 20)$ . The maximum of the objective function $Z = 10 x + 20 y$ is:

1 600

2 550

3 400

4 450

Explanation:

(D) :
| Corner points | $Z = 10 x + 20 y$ |
| :--- | :--- |
| $(0, 10)$ | $0 + 200 = 200$ |
| $(5, 5)$ | $10 \times 5 + 20 \times 5 = 150$ |
| $(15, 15)$ | $10 \times 15 + 20 \times 15 =$ |
| | 450 |
| $(0, 20)$ | $0 + 400 = 400$ |
Maximum of the objective function, $x = 450$

GUJCET-2023

Linear Inequalities and Linear Programming

88613 If the number of available constraints is 3 and the number of parameters to be optimized is 4 , then

1 The objective function can be optimized

2 The constraint are short in number

3 The solution is problem oriented

4 None of these

BITSAT-2019

Linear Inequalities and Linear Programming

88622 Consider a Linear Programming Problem: Minimize $Z = 5 x + 3 y$ , Subject to : $3 x + y \geq 10$ , $2 x + 2 y \geq 14$ and $x + 2 y \geq 9$ . Which one of the following points lies outside the feasible region?

1

(1, 9)

2

(4, 2)

3

(6, 2)

4

(12, 2)

Explanation:

(B) : Given,
Minimize, $Z = 5 x + 3 y$
Subject to
$3 x + y \geq 10$
$2 x + 2 y \geq 14$
And $x + 2 y \geq 9$
On plotting the constraints, we get-
$A (9, 0), B (5, 2), C (\frac{3}{2}, \frac{11}{2})$ and $D (0, 10)$ as corner
point of feasible region which is unbounded.
Now, $(1, 9), (6, 2)$ and $(12, 2)$ lies inside the feasible regions and point $(4, 2)$ lies outside the feasible region.

J&K CET-2019

Linear Inequalities and Linear Programming

88623 The corner points of the feasible region determined by the system of linear constraints are $(0, 10), (5, 5), (25, 20)$ and $(0, 30)$ . Let $Z =$ $p x + q y$ , where $p, q > 0$ , Condition on $p$ and $q$ so that the maximum of $Z$ occurs at both the points $(25, 20)$ and $(0, 30)$ is

1

5 p = 2 q

2

2 p = 5 q

3

p = 2 q

4

q = 3 p

Explanation:

(A) : Given,
Let, $z = px + qy$ , where $p, q > 0$
And given maximum of $Z$ occurs at both the points (25, $20)$ and $(0, 30)$ .
Then
${Max}_{(25, 20)} = {MaxZ}_{(0, 30)}$
$25 p + 20 q = 0 + 30 q$
$25 p = 10 q$
Therefore, condition on $p$ and $q$
$5 p = 2 q$

GUJCET-2017

Linear Inequalities and Linear Programming

88624 The coordinates of the corner points of the bounded feasible region are $(0, 10), (5, 5), (15$ , $15)$ and $(0, 20)$ . The maximum of the objective function $Z = 10 x + 20 y$ is:

1 600

2 550

3 400

4 450

Explanation:

(D) :
| Corner points | $Z = 10 x + 20 y$ |
| :--- | :--- |
| $(0, 10)$ | $0 + 200 = 200$ |
| $(5, 5)$ | $10 \times 5 + 20 \times 5 = 150$ |
| $(15, 15)$ | $10 \times 15 + 20 \times 15 =$ |
| | 450 |
| $(0, 20)$ | $0 + 400 = 400$ |
Maximum of the objective function, $x = 450$

GUJCET-2023

Linear Inequalities and Linear Programming

88613 If the number of available constraints is 3 and the number of parameters to be optimized is 4 , then

1 The objective function can be optimized

2 The constraint are short in number

3 The solution is problem oriented

4 None of these

BITSAT-2019

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