QBManager

Linear Inequalities and Linear Programming

88552 If $Z = 10 x + 25 y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x + y \leq 5, x \geq 0, y \geq 0$ then $Z$ is maximum at the point
is maximum at the point

1

(2, 3)

2

(2, 4)

3

(1, 6)

4

(4, 3)

Explanation:

(A) : Given,
$Z = 10 x + 25 y$
Subject to $0 \leq x \leq 3$
$0 \leq y \leq 3$
$x + y \leq 5$
$x, y \geq 0$
Using graphical method we get

Solve $x = 3$ and $x + y = 5$ we get $x = 3, y = 2$
We solve to get corner points $A, B$ and $C$ as follows.
For point $A$ solve $y = 3$ and $x + y = 5$
We get $x = 2, y = 3$
Now calculate value to $z$ as follows:
At Point A $(2, 3) Z = 20 + 75 = 95$
At Point B $(3, 3), Z = 30 + 75 = 105$
At Point $C (3, 2), Z = 30 + 50 = 80$
At Point $D = (3, 0), Z = 30$
At Point $E (0, 3), Z = 75$
So the max value of $Z$ is 95 which is at point $(2, 3)$ .

MHT CET-2020

Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize $z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq 0$ has

1 No solution

2 One solution

3 Two solutions

4 Infinite solutions

Explanation:

(A) : Given,
$z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq$ 0
Using graphical method, we get -

So, we do not get any $x + y = 1$
Corner point so, the system has no solution.

MHT CET-2020

Linear Inequalities and Linear Programming

88554 The objective function $z = 4 x_{1} + 5 x_{2}$ , subject to $2 x_{1} + x_{2} \geq 7, 2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0$ has minimum value at the point

1 On X - axis

2 On Y - axis

3 At the origin

4 On the use parallel to

x

-axis

Explanation:

(A) : Given, $z = 4 x_{1} + 5 x_{2}$ subject to $2 x_{1} + x_{2} \geq 7$ , $2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3$ and $x_{1}, x_{2} \geq 0$ Using graphical method, we get -

Let as calculate point $A$ solve $x_{2} = 3$ and $2 x_{1} + x_{2} = 7$ we get point $A$ as $x_{1} = 2$ The point $A$ is $(2, 3)$
Similarly we find point $B$ as follows
Solve $x_{2} = 3$ and $2 x_{1} + 3 x_{2} = 15$
We get $2 x_{1} + 9 = 15$
$2 x_{1} = 6$
$x_{1} = 3$
$∴$ The point $B$ is $(3, 3)$
Point $(\frac{15}{2}, 0)$ is point $C$ .
and Point D is $(\frac{7}{2}, 0)$
Now, at Point $A (2, 3) z = 4 x_{1} + 5 x_{2}$
at Point B $(3, 3), z = 12 + 15 = 27$
at Point $C (\frac{15}{2}, 0), z = 4 \times \frac{15}{2} + 0 = 30$
at Point $D (\frac{7}{2}, 0), z = 4 \times \frac{7}{2} + 0 = 14$
$∴$ minimum value is 14 which is point $P$ , which is on the $X$ axis.

MHT CET-2017

Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1

5 x + 4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0

2

5 x + 4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

3

5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

4

5 x + 4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0

Explanation:

(C) : In the given figure $x \leq 6, y \leq 3$ and
Let us find the equation of line passing through Point $(0, 5)$ and $(4, 0)$ $y - 5 = \frac{0 - 5}{4 - 0} (x - 0)$
$y - 5 = \frac{- 5}{4} x$
$4 y - 20 = - 5 x$
or $5 x + 4 y = 20$
Therefore, $5 x + 4 y \geq 20$
So, we get $5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$

Karnataka CET-2021

Linear Inequalities and Linear Programming

88552 If $Z = 10 x + 25 y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x + y \leq 5, x \geq 0, y \geq 0$ then $Z$ is maximum at the point
is maximum at the point

1

(2, 3)

2

(2, 4)

3

(1, 6)

4

(4, 3)

Explanation:

(A) : Given,
$Z = 10 x + 25 y$
Subject to $0 \leq x \leq 3$
$0 \leq y \leq 3$
$x + y \leq 5$
$x, y \geq 0$
Using graphical method we get

Solve $x = 3$ and $x + y = 5$ we get $x = 3, y = 2$
We solve to get corner points $A, B$ and $C$ as follows.
For point $A$ solve $y = 3$ and $x + y = 5$
We get $x = 2, y = 3$
Now calculate value to $z$ as follows:
At Point A $(2, 3) Z = 20 + 75 = 95$
At Point B $(3, 3), Z = 30 + 75 = 105$
At Point $C (3, 2), Z = 30 + 50 = 80$
At Point $D = (3, 0), Z = 30$
At Point $E (0, 3), Z = 75$
So the max value of $Z$ is 95 which is at point $(2, 3)$ .

MHT CET-2020

Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize $z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq 0$ has

1 No solution

2 One solution

3 Two solutions

4 Infinite solutions

Explanation:

(A) : Given,
$z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq$ 0
Using graphical method, we get -

So, we do not get any $x + y = 1$
Corner point so, the system has no solution.

MHT CET-2020

Linear Inequalities and Linear Programming

88554 The objective function $z = 4 x_{1} + 5 x_{2}$ , subject to $2 x_{1} + x_{2} \geq 7, 2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0$ has minimum value at the point

1 On X - axis

2 On Y - axis

3 At the origin

4 On the use parallel to

x

-axis

Explanation:

(A) : Given, $z = 4 x_{1} + 5 x_{2}$ subject to $2 x_{1} + x_{2} \geq 7$ , $2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3$ and $x_{1}, x_{2} \geq 0$ Using graphical method, we get -

Let as calculate point $A$ solve $x_{2} = 3$ and $2 x_{1} + x_{2} = 7$ we get point $A$ as $x_{1} = 2$ The point $A$ is $(2, 3)$
Similarly we find point $B$ as follows
Solve $x_{2} = 3$ and $2 x_{1} + 3 x_{2} = 15$
We get $2 x_{1} + 9 = 15$
$2 x_{1} = 6$
$x_{1} = 3$
$∴$ The point $B$ is $(3, 3)$
Point $(\frac{15}{2}, 0)$ is point $C$ .
and Point D is $(\frac{7}{2}, 0)$
Now, at Point $A (2, 3) z = 4 x_{1} + 5 x_{2}$
at Point B $(3, 3), z = 12 + 15 = 27$
at Point $C (\frac{15}{2}, 0), z = 4 \times \frac{15}{2} + 0 = 30$
at Point $D (\frac{7}{2}, 0), z = 4 \times \frac{7}{2} + 0 = 14$
$∴$ minimum value is 14 which is point $P$ , which is on the $X$ axis.

MHT CET-2017

Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1

5 x + 4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0

2

5 x + 4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

3

5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

4

5 x + 4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0

Explanation:

(C) : In the given figure $x \leq 6, y \leq 3$ and
Let us find the equation of line passing through Point $(0, 5)$ and $(4, 0)$ $y - 5 = \frac{0 - 5}{4 - 0} (x - 0)$
$y - 5 = \frac{- 5}{4} x$
$4 y - 20 = - 5 x$
or $5 x + 4 y = 20$
Therefore, $5 x + 4 y \geq 20$
So, we get $5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$

Karnataka CET-2021

Linear Inequalities and Linear Programming

88552 If $Z = 10 x + 25 y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x + y \leq 5, x \geq 0, y \geq 0$ then $Z$ is maximum at the point
is maximum at the point

1

(2, 3)

2

(2, 4)

3

(1, 6)

4

(4, 3)

Explanation:

(A) : Given,
$Z = 10 x + 25 y$
Subject to $0 \leq x \leq 3$
$0 \leq y \leq 3$
$x + y \leq 5$
$x, y \geq 0$
Using graphical method we get

Solve $x = 3$ and $x + y = 5$ we get $x = 3, y = 2$
We solve to get corner points $A, B$ and $C$ as follows.
For point $A$ solve $y = 3$ and $x + y = 5$
We get $x = 2, y = 3$
Now calculate value to $z$ as follows:
At Point A $(2, 3) Z = 20 + 75 = 95$
At Point B $(3, 3), Z = 30 + 75 = 105$
At Point $C (3, 2), Z = 30 + 50 = 80$
At Point $D = (3, 0), Z = 30$
At Point $E (0, 3), Z = 75$
So the max value of $Z$ is 95 which is at point $(2, 3)$ .

MHT CET-2020

Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize $z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq 0$ has

1 No solution

2 One solution

3 Two solutions

4 Infinite solutions

Explanation:

(A) : Given,
$z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq$ 0
Using graphical method, we get -

So, we do not get any $x + y = 1$
Corner point so, the system has no solution.

MHT CET-2020

Linear Inequalities and Linear Programming

88554 The objective function $z = 4 x_{1} + 5 x_{2}$ , subject to $2 x_{1} + x_{2} \geq 7, 2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0$ has minimum value at the point

1 On X - axis

2 On Y - axis

3 At the origin

4 On the use parallel to

x

-axis

Explanation:

(A) : Given, $z = 4 x_{1} + 5 x_{2}$ subject to $2 x_{1} + x_{2} \geq 7$ , $2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3$ and $x_{1}, x_{2} \geq 0$ Using graphical method, we get -

Let as calculate point $A$ solve $x_{2} = 3$ and $2 x_{1} + x_{2} = 7$ we get point $A$ as $x_{1} = 2$ The point $A$ is $(2, 3)$
Similarly we find point $B$ as follows
Solve $x_{2} = 3$ and $2 x_{1} + 3 x_{2} = 15$
We get $2 x_{1} + 9 = 15$
$2 x_{1} = 6$
$x_{1} = 3$
$∴$ The point $B$ is $(3, 3)$
Point $(\frac{15}{2}, 0)$ is point $C$ .
and Point D is $(\frac{7}{2}, 0)$
Now, at Point $A (2, 3) z = 4 x_{1} + 5 x_{2}$
at Point B $(3, 3), z = 12 + 15 = 27$
at Point $C (\frac{15}{2}, 0), z = 4 \times \frac{15}{2} + 0 = 30$
at Point $D (\frac{7}{2}, 0), z = 4 \times \frac{7}{2} + 0 = 14$
$∴$ minimum value is 14 which is point $P$ , which is on the $X$ axis.

MHT CET-2017

Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1

5 x + 4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0

2

5 x + 4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

3

5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

4

5 x + 4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0

Explanation:

(C) : In the given figure $x \leq 6, y \leq 3$ and
Let us find the equation of line passing through Point $(0, 5)$ and $(4, 0)$ $y - 5 = \frac{0 - 5}{4 - 0} (x - 0)$
$y - 5 = \frac{- 5}{4} x$
$4 y - 20 = - 5 x$
or $5 x + 4 y = 20$
Therefore, $5 x + 4 y \geq 20$
So, we get $5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$

Karnataka CET-2021

Linear Inequalities and Linear Programming

88552 If $Z = 10 x + 25 y$ subject to $0 \leq x \leq 3, 0 \leq y \leq 3, x + y \leq 5, x \geq 0, y \geq 0$ then $Z$ is maximum at the point
is maximum at the point

1

(2, 3)

2

(2, 4)

3

(1, 6)

4

(4, 3)

Explanation:

(A) : Given,
$Z = 10 x + 25 y$
Subject to $0 \leq x \leq 3$
$0 \leq y \leq 3$
$x + y \leq 5$
$x, y \geq 0$
Using graphical method we get

Solve $x = 3$ and $x + y = 5$ we get $x = 3, y = 2$
We solve to get corner points $A, B$ and $C$ as follows.
For point $A$ solve $y = 3$ and $x + y = 5$
We get $x = 2, y = 3$
Now calculate value to $z$ as follows:
At Point A $(2, 3) Z = 20 + 75 = 95$
At Point B $(3, 3), Z = 30 + 75 = 105$
At Point $C (3, 2), Z = 30 + 50 = 80$
At Point $D = (3, 0), Z = 30$
At Point $E (0, 3), Z = 75$
So the max value of $Z$ is 95 which is at point $(2, 3)$ .

MHT CET-2020

Linear Inequalities and Linear Programming

88553 The L.P.P. to maximize $z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq 0$ has

1 No solution

2 One solution

3 Two solutions

4 Infinite solutions

Explanation:

(A) : Given,
$z = x + y$ subject to $x + y \leq 1, 2 x + 2 y \geq 6, x \geq 0, y \geq$ 0
Using graphical method, we get -

So, we do not get any $x + y = 1$
Corner point so, the system has no solution.

MHT CET-2020

Linear Inequalities and Linear Programming

88554 The objective function $z = 4 x_{1} + 5 x_{2}$ , subject to $2 x_{1} + x_{2} \geq 7, 2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3, x_{1}, x_{2} \geq 0$ has minimum value at the point

1 On X - axis

2 On Y - axis

3 At the origin

4 On the use parallel to

x

-axis

Explanation:

(A) : Given, $z = 4 x_{1} + 5 x_{2}$ subject to $2 x_{1} + x_{2} \geq 7$ , $2 x_{1} + 3 x_{2} \leq 15, x_{2} \leq 3$ and $x_{1}, x_{2} \geq 0$ Using graphical method, we get -

Let as calculate point $A$ solve $x_{2} = 3$ and $2 x_{1} + x_{2} = 7$ we get point $A$ as $x_{1} = 2$ The point $A$ is $(2, 3)$
Similarly we find point $B$ as follows
Solve $x_{2} = 3$ and $2 x_{1} + 3 x_{2} = 15$
We get $2 x_{1} + 9 = 15$
$2 x_{1} = 6$
$x_{1} = 3$
$∴$ The point $B$ is $(3, 3)$
Point $(\frac{15}{2}, 0)$ is point $C$ .
and Point D is $(\frac{7}{2}, 0)$
Now, at Point $A (2, 3) z = 4 x_{1} + 5 x_{2}$
at Point B $(3, 3), z = 12 + 15 = 27$
at Point $C (\frac{15}{2}, 0), z = 4 \times \frac{15}{2} + 0 = 30$
at Point $D (\frac{7}{2}, 0), z = 4 \times \frac{7}{2} + 0 = 14$
$∴$ minimum value is 14 which is point $P$ , which is on the $X$ axis.

MHT CET-2017

Linear Inequalities and Linear Programming

88555 The shaded region is the solution set of the inequalities.

1

5 x + 4 y \geq 20, x \leq 6, y \geq 3, x \geq 0, y \geq 0

2

5 x + 4 y \leq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

3

5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0

4

5 x + 4 y \geq 20, x \geq 6, y \leq 3, x \geq 0, y \geq 0

Explanation:

(C) : In the given figure $x \leq 6, y \leq 3$ and
Let us find the equation of line passing through Point $(0, 5)$ and $(4, 0)$ $y - 5 = \frac{0 - 5}{4 - 0} (x - 0)$
$y - 5 = \frac{- 5}{4} x$
$4 y - 20 = - 5 x$
or $5 x + 4 y = 20$
Therefore, $5 x + 4 y \geq 20$
So, we get $5 x + 4 y \geq 20, x \leq 6, y \leq 3, x \geq 0, y \geq 0$

Karnataka CET-2021

Question Bank Series

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