Linear Programming Test

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Linear Programming Test - 29

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Question 1
1 / -0

If $$\displaystyle x\ge 0$$
$$\displaystyle 3y-2x\ge -12$$
$$\displaystyle 2x+5y\le 20$$
The area of the triangle formed in the xy plane by the system of inequalities above is:

A
$$60$$

B
$$30$$

C
$$40$$

D
$$50$$

Solution

given, $$x\geq 0$$
$$3y-2x \geq -12$$ and $$2x+5y\leq 20$$

first, draw the graph for equations $$x= 0$$
$$3y-2x = -12$$ and $$2x+5y = 20$$

$$x=0$$ is Y-axis. Hence $$x\geq 0$$ includes the above region of the line.

for $$3y-2x = -12$$
substitute y=0 we get, $$-2x=-12 \implies x=6$$
substitute x=0 we get, $$3y=-12 \implies y=-4$$
therefore, $$3y-2x = -12$$ line passes through (6,0) and (0,-4) as shown in fig.
Hence, $$3y-2x \geq -12$$includes the region above the line.

similarly for $$2x+5y = 20$$
substitute y=0 we get, $$2x=20 \implies x=10$$
substitute x=0 we get, $$5y=20 \implies y=4$$
therefore, $$2x+5y = 20$$ line passes through (10,0) and (0,4) as shown in fig.
Hence, $$2x+5y\leq 20$$includes the region below the line

the shaded region as shown in figure is intersection region
distance between (0,4) and (0,-4) is 8 units
adding $$3y-2x = -12$$ and $$2x+5y = 20 \implies y=1$$
substituting y=1 in any one of equation gives $$x=7.5$$

therefore, shaded triangle base =8 units and
height = 7.5 unit
Area = $$\dfrac{1}{2}\times base\times height$$
$$= \dfrac{1}{2}\times 8\times 7.5 = 30$$
Question 2
1 / -0

Conclude from the following:
$$n^2 > 10$$, and n is a positive integer.
A: $$n^3$$
B: $$50$$

A
The quantity A is may be greater or smaller than B.

B
The quantity B is greater than A.

C
The two quantities are equal.

D
The relationship cannot be determined from the information given.

Solution

given, $$n^2 > 10$$ and $$n >0 $$
multiplying both equations we get
$$n^3>0$$
so, it may be greater than or less than 50
Hence, quantity A is may be greater or smaller than B
Question 3
1 / -0

Minimize : $$z=3x+y$$, subject to $$2x+3y\le 6, x+y \ge 1, x\ge 0, y\ge 0$$

A
$$x=1, y=1$$

B
$$x=0, y=1$$

C
$$x=1, y=0$$

D
$$x=-1, y=-1$$

Solution

Question 4

1 / -0

A manufacturer considers that men and women workers are equally efficient and so he pays them at the same rate. He has $$30$$ workers (male and female) and $$17$$ units capital; which he uses to produce two types of goods $$A$$ and $$B$$. To produce one unit of $$A$$, $$2$$ workers and $$3$$ units of capital are required while $$3$$ workers and $$1$$ unit of capital is required to produce one unit of $$B$$. If $$A$$ and $$B$$ are priced at $$Rs.\ 100$$ and $$Rs.\ 120$$ per unit respectively, how should he use his resources to maximise the total revenue? Form the LPP and solve graphically.
Do you agree with this view of the manufacturer that men and women workers are equally efficient and so should be paid at the same rate?

$$1260$$

$$1130$$

$$1290$$

$$3421$$

Solution

Let the number of goods of type A and B produced be respectively x and y.

To maximize, $$Z=(100x+120y)$$

Subject to the constraints:

$$2x+3y\leq30$$ ---- (1)

$$3x+y\leq17$$---- (2) where $$x, y\leq 0.$$

Take the testing points as $$(0, 0)$$ for (1) we have: $$2(0) + 3(0) \leq 30\Rightarrow0\leq 30$$, which is true.

Take the testing points as $$(0, 0)$$ for (2) we have: $$3(0) + (0) \leq 17\Rightarrow0\leq 17$$, which is true.

The shaded region $$OACBO$$ as shown in the given figure is the feasible region, which is bounded.

The coordinates of the corner points of the feasible region are $$A(\frac{17}{3},0), E(3,8), C(0, 10) and O(0, 0).$$

So, Value of Z at $$A\left(\dfrac{17}{3}, 0\right)=\dfrac{1700}{3}$$

Value of Z at $$B(0, 10)=1200$$

Value of Z at $$C(3, 8)=1260$$

Value of Z at $$O(0, 0)=0$$

Pts.	Coordinate	$$Z^{max} = 100x + 120y$$
$$O$$	$$(0,0)$$	$$Z=0$$
$$A$$	$$\left(\dfrac{17}3,0\right)$$	$$Z = \dfrac{1700}3$$
$$E$$	$$(3,8)$$	$$Z = 300+960 = 1260$$
$$C$$	$$(0,10)$$	$$Z = 1200$$

The maximum value of Z is $$Rs.1260$$ which occurs at $$x = 3$$ and $$y = 8.$$

Thus the factory must produce $$3$$ units and $$8$$ units of the goods of type A and B respectively. The maximum obtained profit earned by the factory by producing these items is $$Rs. 1260.$$

Yes, we agree with the view of manufacturer that men and women workers are equally efficient and so should be paid at the same rate.

Question 5
1 / -0

If $$x+y \leq 2, x\leq 0, y\leq 0$$ the point at which maximum value of $$3x+2y$$ attained will be.

A
$$(0, 0)$$

B
$$\left ( \dfrac{1}{2}, \dfrac{1}{2} \right )$$

C
$$(0, 2)$$

D
$$(2, 0)$$

Solution

$$x \le 0$$ and $$y \le 0$$ represents third Quadrant
$$x+y \le 2$$ represents the region below the line $$x+y \le 2$$ $$($$ the region which contains origin$$)$$
The common region of given set of equations is third quadrant $$($$ including negative $$x$$ axis and negative $$y$$ axis$$)$$
Since $$x$$ and $$y$$ values are $$\le 0$$ in the third quadrant , the maximum value of $$3x+2y$$ occurs at $$x=0$$ and $$y=0$$ and the maximum value is $$0$$
Therefore the correct option is $$A$$
Question 6
1 / -0

In figure 32, the shaded region within the triangle is the intersection of the sets of ordered pairs described by which of the following inequalities?

A
y < x, x < 2

B
y < 2x, x < 2

C
y < 2x, x < 2, x > 0

D
y < 2x, y < 2, x > 0

E
y < 2x, x < 2, y > 0

Solution
Question 7
1 / -0

The linear programming problem:
Maximize $$z={ x }_{ 1 }+{ x }_{ 2 }$$
Subject to constraints
$$\quad { x }_{ 1 }+2{ x }_{ 2 }\le 2000,{ x }_{ 1 }+{ x }_{ 2 }\le 1500,\quad { x }_{ 2 }\le 600,\quad { x }_{ 1 }\ge 0$$

A
No feasible solution

B
Unique optimal solution

C
A finite number of optimal solutions

D
Infinite number of optimal solutions

Solution
Question 8
1 / -0

Solve the following LPP graphically. Maximize or minimize $$Z = 3x + 5y$$ subject to
$$3x - 4y \geq -12$$
$$2x - y + 2\geq 0$$
$$2x + 3y - 12\geq 0$$
$$0 \leq x \leq 4$$
$$y \geq 2$$.

A
Min. value $$19$$ at $$(5, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.

B
Min. value $$30$$ at $$(3, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.

C
Min. value $$19$$ at $$(3, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.

D
Min. value $$8$$ at $$(3, 2)$$ and Max. value $$42$$ at $$(4, 6)$$.

Solution

To maximise : $$3x+5y$$
Constraints:
1) $$3x-4y\ge -12$$
2) $$2x-y\ge -2$$
3) $$2x+3y\ge 12$$
$$0\le x\le 4$$
$$y\ge 2$$

The shaded region satisfies the given constraints.
From critical points;
Minimum at $$(3,2)$$
$$Z=3(3)+5(2)=9+10=19$$

Maximum at $$(4,6)$$
$$Z=4(3)+5(6)=42$$
To sum up,
min. $$19$$ at $$(3,2)$$ and
max. $$42$$ at $$(4,6)$$
Question 9
1 / -0

Let $$P(-1, 0), Q(0, 0)$$ and $$R(3, 3\sqrt{3})$$ be three points. The equation of the bisector of the angle PQR is?

A
$$x+\sqrt{3}y=0$$

B
$$\sqrt{3}x+y=0$$

C
$$x+\dfrac{\sqrt{3}}{2}y=0$$

D
$$\dfrac{\sqrt{3}}{2}x+y=0$$

Solution

Given points are $$P(-1,0),Q(0,0)$$ and $$R(3, 3\sqrt{3})$$
Slope of $$l\, ne \,QR=\sqrt 3$$
Slope of lime $$QM=\tan \dfrac{2\pi}{3}$$
$$m=(\sqrt 3)$$
Hence, eqn of line $$QM$$ is $$y=mx+c$$
$$\therefore y=0\, -\sqrt {3x} +0$$ [taking $$Q(0,0)$$]
$$\boxed {\sqrt{3x}+y=0}$$
Question 10
1 / -0

Use graph paper for this question:

A
Plot the points $$A(-4, 2)$$ and $$B(2, 4)$$

B
$${A}^{1}$$ is the image of $$A$$ when reflected in the line $$x=0$$. Write the co-ordinates of $${A}^{1}$$.

C
$${B}^{1}$$ is the image of $$B$$ when reflected in the line $$A{A}^{1}$$. Write the co-ordinates of $${B}^{1}$$.

D
Write the geometrical name of the figure $$AB{A}^{1}{B}^{1}$$

Solution

1) plot points $$A(-4,2) and\, B(2,4)$$
2) A' image of A in $$x=0$$ $$\therefore A'\equiv (4,2)$$
3) B' image of B in AA' $$\therefore B'=(2,0)$$
4) shape of ABA'B' is shaped quadrilateral.