Linear Programming Test

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Linear Programming Test - 25

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Question 1
1 / -0

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

A
On x-axis

B
On y-axis

C
At the origin

D
On the parallel to x-axis

Solution

Value of $$z = 4x_{1} + 5x_{2}$$
Convert the given inequalities into equalities to get the corner points
$$2x_{1} + x_{2} = 7$$ ....... $$(i)$$
At $$x_1=0, x_{2}=7$$ and $$x_2=0, x_{1}=3.5$$
So, the corner points of $$(i)$$ are $$(0,7)$$ and $$(3.5,0)$$
$$2x_{1} + 3x_{2} = 15$$ ...... $$(ii)$$
At $$x_1=0, x_{2}=5$$ and $$x_{2}=0, x_{1}=7.5$$
So, the corner points of $$(i)$$ are $$(0,5)$$ and $$(7.5,0)$$
$$x_{2}=3$$ ...... $$(iii)$$
Plot these corner points on the graph paper and the line given in $$(iii)$$
The shaded part shows the feasible region.
At $$x_{2}=3, x_1=2$$ in $$(i)$$ and $$x_{1}=3$$ in $$(ii)$$
The corner points of the feasible region are $$(3.5,0), (7.5,0), (3,3)$$ and $$(2,3)$$
Corner points $$Z=4x_{2}+5x_{2}$$
$$(3.5,0)$$ $$14$$
$$(7.5,0)$$ $$30$$
$$(3,3)$$ $$27$$
$$(2,3)$$ $$23$$

Minimum value of $$Z= 14$$ and it lies on $$x$$-axis.
Question 2
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The objective function of LPP defined over the convex set attains its optimum value at

A
atleast two of the corner points

B
all the corner points

C
atleast one of the corner points

D
none of the corner points

Solution

Let $$Z=ax+by$$ be the objective function

When $$Z$$ has optimum value(maximum or minimum), where the variables

$$x$$ and $$y$$ are subject to constraints described by linear inequalities, this optimum value must occur at a corner points of the feasible region.

Thus, the function attains its optimum value at one of the corner points.

Hence, C is correct.
Question 3
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The constraints
$$-{ x }_{ 1 }+{ x }_{ 2 }\le 1$$
$$-{ x }_{ 1 }+3{ x }_{ 2 }\le 9$$
$${x}_{1},{x}_{2}\ge 0$$ defines on

A
Bounded feasible space

B
Unbounded feasible space

C
Both bounded and unbounded feasible

D
None of the above

Solution

Draw the following curves and find common area, which will give you feasible solutions.
So, our constraints are:
$$-{ x }_{ 1 }+{ x }_{ 2 }\le 1$$ -------------(i)
$$-{ x }_{ 1 }+3{ x }_{ 2 }\le 9$$ ----------(ii)
$${ x }_{ 1 }{ x }_{ 2 }\ge 0$$ -------------(iii)

We can see from the graph clearly that the constraints have unbounded feasible space
Question 4
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An article manufactured by a company consists of two parts $$X$$ and $$Y$$. In the process of manufacture of the part $$X$$. $$9$$ out of $$100$$ parts may be defective. Similarly $$5$$ out of $$100$$ are likely to be defective in part $$Y$$. Calculate the probability that the assembled product will not be defective.

A
$$0.86$$

B
$$0.864$$

C
$$0.8456$$

D
$$0.8645$$

Solution

Let $$A=$$ Part $$X$$ is not defective

Probability of $$A$$ is $$P(A)=\dfrac {91}{100}$$

$$B=$$Part $$Y$$ is not defective.

Probability of $$B$$ is $$P(B)=\dfrac {95}{100}$$

Required probability $$=P(A\cap B)=P(A)P(B)=\dfrac {91}{100}\times \dfrac {95}{100}=\dfrac{8645}{10000}$$
Question 5
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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 __________.

A
$$q=2p$$

B
$$p=2p$$

C
$$p=q$$

D
$$q=3p$$

Solution

Let $$z_{0}$$ be the maximum value of $$z$$ in the feasible region.
Since maximum occurs at both $$(15,15)$$ and $$(0,20)$$$, the value $$z_{0}$$ is attained at both $$(15,15)$$ and $$(0,20)$$.
$$\implies z_{0}=p(15)+q(15)$$ and $$ z_{0}=p(0)+q(20)$$
$$\implies p(15)+q(15)=p(0)+q(20)$$
$$\implies 15p=5q$$
$$\implies 3p=q$$
Hence, the answer is option (D).
Question 6
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Corner points of the bounded feasible region for an LP problem are $$A(0,5) B(0,3) C(1,0) D(6,0)$$. Let $$z = -50x + 20y$$ be the objective function. Minimum value of z occurs at ______ center point.

A
$$(0,5)$$

B
$$(1,0)$$

C
$$(6,0)$$

D
$$(0,3)$$

Solution

We check the value of the $$z$$ at each of the corner points.
At $$A(0,5)-$$
$$z=-50x+20y=-50(0)+20(5)=100$$
At $$B(0,3)-$$
$$z=-50x+20y=-50(0)+20(3)=60$$
At $$C(1,0)-$$
$$z=-50x+20y=-50(1)+20(0)=-50$$
At $$D(6,0)-$$
$$z=-50x+20y=-50(6)+20(0)=-300$$
Hence, we see that $$z$$ is minimum at $$D(6,0)$$ and minimum value is $$-300$$.
Question 7
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The corner points of the feasible region are $$A(0,0),B(16,0),C(8,16)$$ and $$D(0,24)$$. The minimum value of the objective function $$z=300x+190y$$ is _______

A
$$5440$$

B
$$4800$$

C
$$4560$$

D
$$0$$

Solution

We know that, for a cartesian polygon , the maximum value occurs at the corner points or vertices of the polygon.
Given $$z=300x+190y$$
By substituting $$A(0,0)$$ in the equation we get $$z=0$$
By substituting $$B(16,0)$$ in the equation we get $$z=4800$$
By substituting $$C(8,16)$$ in the equation we get $$z=5440$$
By substituting $$D(0,24)$$ in the equation we get $$z=4560$$
Hence the minimum value of Z occured at $$C(0,0)$$ with $$z=0$$
Question 8
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The given table shows the number of cars manufactured in four different colours on a particular day. Study it carefully and answer the question.

Colour Number of cars manufactured
Vento Creta WagonR
Red 65 88 93
White 54 42 80
Black 66 52 88
Sliver 37 49 74
Which car was twice the number of silver Vento?

A
Silver WagonR

B
Red WagonR

C
Red Vento

D
White Creta

Solution

The number of silver Vento car $$=37$$ (from the table)

Twice the number of silver Vento cars$$ = 2 \times 37=74$$

Now from table we can see that silver WagonR is only car type having $$74 $$ cars
Question 9
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A firm manufactures three products $$A,B$$ and $$C$$. Time to manufacture product $$A$$ is twice that for $$B$$ and thrice that for $$C$$ and if the entire labour is engaged in making product $$A,1600$$ units of this product can be produced.These products are to be produced in the ratio $$3:4:5.$$ There is demand for at least $$300,250$$ and $$200$$ units of products $$A,B$$ and $$C$$ and the profit earned per unit is Rs.$$90,$$ Rs$$40$$ and Rs.$$30$$ respectively.
Raw
material Requirement per unit product(Kg)
A Requirement per unit product(Kg)
B Requirement per unit product(Kg)
C Total availability (kg)
$$P$$ $$6$$ $$5$$ $$2$$ $$5,000$$
$$Q$$ $$4$$ $$7$$ $$3$$ $$6,000$$
Formulate the problem as a linear programming problem and find all the constraints for the above product mix problem.

A
$$3{x}_{1}-4{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$

B
$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$

C
$$4{x}_{1}-3{x}_{2}=0$$ and $$4{x}_{2}-5{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$

D
$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\le0$$

Solution

Formulation of L.P Model
Let $${x}_{1},{x}_{2}$$ and $${x}_{3}$$ denote the number of units of products $$A,B$$ and $$C$$ to be manufactured .
Objective is to maximize the profit.
i.e., maximize $$Z=90{x}_{1}+40{x}_{2}+30{x}_{3}$$
Constraints can be formulated as follows:
For raw material $$P, 6{x}_{1}+5{x}_{2}+2{x}_{3}\le5,000$$
For raw material $$Q, 4{x}_{1}+7{x}_{2}+3{x}_{3}\le6,000$$
Product $$B$$ requires $$\frac{1}{2}$$ and product $$C$$ requires $${\left(\frac{1}{3}\right)}^{rd}$$ the time required for product $$A.$$
Then $$\frac{t}{2}$$ and $$\frac{t}{3}$$ are the times in hours to produce $$B$$ and $$C$$ and since $$1,600$$ units of $$A$$ will need time $$1,600t$$ hours, we get the constraint,
$$t{x}_{1}+\frac{t}{2}{x}_{2}+\frac{t}{3}{x}_{3}\le 1,600t$$ or
$${x}_{1}+\frac{{x}_{2}}{2}+\frac{{x}_{3}}{3}\le1,600$$ or
$$6{x}_{1}+3{x}_{2}+2{x}_{3}\le9,600$$
Market demand requires
$${x}_{1}\ge300, {x}_{2}\ge250,$$ and $${x}_{3}\ge200$$
Finally, since products $$A,B$$ and $$C$$ are to be produced in the ratio $$3:4:5,$$
$${x}_{1}:{x}_{2}:{x}_{3}::3:4:5$$
or $$\frac{{x}_{1}}{3}=\frac{{x}_{2}}{4},$$
and $$\frac{{x}_{2}}{4}=\frac{{x}_{3}}{5}.$$
Thus, there are two additional constraints
$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$
Question 10
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Equation of normal drawn to the graph of the function defined as $$f(x)=\dfrac{\sin x^2}{x}$$, $$x\neq 0$$ and $$f(0)=0$$ at the origin is?

A
$$x+y=0$$

B
$$x-y=0$$

C
$$y=0$$

D
$$x=0$$

Solution

let slope of tangent=T, slope of normal=N
$$ T=\cfrac { dy }{ dx } =\cfrac { x.cos{ x }^{ 2 }\times 2x-sin{ x }^{ 2 } }{ { x }^{ 2 } } $$
$$ T\quad at(x=0)=\cfrac { cos{ x }^{ 2 }\times 2{ x }^{ 2 }-sin{ x }^{ 2 } }{ { x }^{ 2 } } $$
$$ x=0\quad \implies\quad 2cos{ x }^{ 2 }-\cfrac { sin{ x }^{ 2 } }{ { x }^{ 2 } } $$
$$ \quad \quad \quad (\cfrac { sinx }{ x } \quad at\quad x=1)$$
$$ \quad \implies 2-1=1\quad $$
$$ \quad \quad N\times T=-1$$
$$ \quad \quad N=-1$$
$$ at\quad x=0,y=0$$
$$ \quad y=mx$$
$$ \quad y=-x$$
$$ \quad x+y=0$$

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Re-Attempt Weekly Quiz Competition

Colour	Number of cars manufactured
Colour	Vento	Creta	WagonR
Red	65	88	93
White	54	42	80
Black	66	52	88
Sliver	37	49	74

Raw material	Requirement per unit product(Kg) A	Requirement per unit product(Kg) B	Requirement per unit product(Kg) C	Total availability (kg)
$$P$$	$$6$$	$$5$$	$$2$$	$$5,000$$
$$Q$$	$$4$$	$$7$$	$$3$$	$$6,000$$

Linear Programming Test - 25

Result

Linear Programming Test - 25

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Rank

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SHARING IS CARING

Question 1

On x-axis

On y-axis

At the origin

On the parallel to x-axis

Solution

Question 2

atleast two of the corner points

all the corner points

atleast one of the corner points

none of the corner points

Solution

Question 3

Bounded feasible space

Unbounded feasible space

Both bounded and unbounded feasible

None of the above

Solution

Question 4

$$0.86$$

$$0.864$$

$$0.8456$$

$$0.8645$$

Solution

Question 5

$$q=2p$$

$$p=2p$$

$$p=q$$

$$q=3p$$

Solution

Question 6

$$(0,5)$$

$$(1,0)$$

$$(6,0)$$

$$(0,3)$$

Solution

Question 7

$$5440$$

$$4800$$

$$4560$$

$$0$$

Solution

Question 8

Silver WagonR

Red WagonR

Red Vento

White Creta

Solution

Question 9

$$3{x}_{1}-4{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$

$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$

$$4{x}_{1}-3{x}_{2}=0$$ and $$4{x}_{2}-5{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\ge0$$

$$4{x}_{1}-3{x}_{2}=0$$ and $$5{x}_{2}-4{x}_{3}=0$$ where $${x}_{1},{x}_{2},{x}_{3}\le0$$

Solution

Question 10

$$x+y=0$$

$$x-y=0$$

$$y=0$$

$$x=0$$

Solution

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