Linear Programming Test

Result

Linear Programming Test - 27

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

Convex set: A set of point in a plane is said to be convex set it the line segment joining any two points in the set, completely lies in the set.
Which of the following sets represent the convex sets ? jistify your answer.

A

B

C

D

E

Solution

Image A-Convex set, $$\because $$ line joining any two points in the set lies inside the set
Image B-convex set
Image C-Not a convex set
Image D-Not a convex set
Image E-Convex set
Image F-Not a convex set
Question 2
1 / -0

The feasible region of an LPP is shown in the figure. If $$z=3x+9y$$, then the minimum value of z occurs at?

A
$$(5, 5)$$

B
$$(0, 10)$$

C
$$(0, 20)$$

D
$$(15, 15)$$

Solution

z+3x+9y.
$$z+3x+9y$$

$$z_{(5.5)}=15+45=60$$

$$z_{(0, 0)}=0+90=90$$

Minimum value at $$(5, 5)$$.
Question 3
1 / -0

For the LPP; maximise $$z=x+4y$$ subject to the constraints $$x+2y\leq 2$$, $$x+2y\geq 8$$, $$x, y\geq 0$$.

A
$$z_{max}=4$$

B
$$z_{max}=8$$

C
$$z_{max}=16$$

D
Has no feasible solution

Solution

$$x+2y\leq 2$$
$$x+2y\geq 8$$
$$x, y\geq 0$$.
Question 4
1 / -0

If $$ 2x - y +1 = 0 $$ is tangent to the hyperbola $$ \frac {x^2}{a^2} - \frac {y^2}{16} = 1 $$ then which of the following CANNOT be sides of a right angled triangle?

A
a,4,1

B
2a,4,1

C
a,4,2

D
2a,8,1

Solution
Question 5
1 / -0

The maximum value of $$z=9x+11y$$ subject to $$3x+2y\leq 12, 2x+3y\leq 12$$, $$x\geq 0, y\geq 0$$ is _________.

A
$$44$$

B
$$54$$

C
$$36$$

D
$$48$$

Solution

$$3x + 2y = 12$$ ....$$(1)$$
$$2x + 3y = 12$$ ....$$(2)$$

Now, Solving eq. $$(1)$$ and $$(2)$$ we get,
$$x = \dfrac{12}{5}, y = \dfrac{12}{5}$$

$$\therefore\ B \left (\dfrac{12}{5}, \dfrac{12}{5} \right ) $$

Now, $$Z = 9x + 11 y$$

(a) at $$(0,0) \Rightarrow\ Z = 9(0) + 11 (0) = 0$$

(b) at $$(0,4) \Rightarrow\ Z = 9(0) + 11(4) = 44$$

(c) at $$(4,0) \Rightarrow\ Z = 9(4) + 11 (0) = 36$$

(d) at $$\left (\dfrac{12}{5}, \dfrac{12}{5} \right ) \Rightarrow\ Z = 9 \left (\dfrac{12}{5} \right) + 11 \left (\dfrac{12}{5} \right ) $$

$$ = \dfrac{12}{5} (9 + 11 ) - \dfrac{12}{5} \times 20$$

$$ = 12 \times 4 $$

$$ = 48$$

$$\therefore\ Z (max) = 48$$
Question 6
1 / -0

The minimum value of $$z=10x+25y$$ subject to $$0\leq x\leq 3, 0\leq y\leq 3, x+y\geq 5$$ is?

A
$$80$$

B
$$95$$

C
$$105$$

D
$$30$$

Solution

From graph it is evident that (x,y) satisfying all required conditions for which z is minimum must lie on x+y=5,
So, for finding minimum we can rewrite z as $$z= 10(5-y)+25y = 15y +50$$
Lower the value of y, lower is value of z . The minimum value of y which satisfy all required condition is 2 which is also evident from graph.
So, minimum value of z is $$15(2)+50=80$$
Question 7
1 / -0

Two towns A and B are 60 km apart. A school is to built to serve 150 students in town A and 50 students in town B. If the total distance to be travelled by all 200 students is to be as small as possible, then the school be built at

A
town B

B
45 km from town A

C
town A

D
45 km from town B

Solution

Let the distance of the school from A be x.
Therefore, the distance of the school from b is 60 - x.
The total distance covered by 200 students is
[150x + 50 (60 - x)] = [100x + 3000]
This is minimum when x = 0.
Hence, the school should be at town A.
Question 8
1 / -0

The feasible solution for a LPP is shown in Fig. Let $$Z = 3x - 4y$$ be the
Objective function, Minimum of Z occurs at

A
(0, 0)

B
(0, 8)

C
(0, 5)

D
(4, 10)

Solution

Hence, the minimum of Z occurs at (0, 8) and its minimum value is -32.
Question 9
1 / -0

The feasible solution for a LPP is shown in Fig. Let Z=3x−4y be the
The objective function, Maximum value of Z + Minimum value of Z is equal to

A
13

B
1

C
-13

D
-17

Solution

maximum value of Z + minimum value of Z.
= 15 - 32 = -17
Question 10
1 / -0

The feasible solution for a LPP is shown in Fig. Let Z=3x−4y be the
Objective function, Maximum of Z occurs at

A
(5, 0)

B
(6, 5)

C
(6, 8)

D
(4, 10)

Solution

Maximum of Z occurs at (5, 0).