Linear Programming Test

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Linear Programming Test - 24

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Question 1
1 / -0

The Convex Polygon Theorem states that the optimum (maximum or minimum) solution of a LPP is attained at atleast one of the ______ of the convex set over which the solution is feasible.

A
origin

B
corner points

C
centre

D
edge

Solution

The fundamental theorem of programming (i.e., Convex Polygon Theorem) states that the optimum value(maximum or minimum) of a linear programming problem over a convex region occur at the corner points.
Question 2
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Directions For Questions

A company manufactures two types of calculators - normal calculator and scientific calculator. Due to demand in the market, the company must produce a minimum of $$100$$ normal and $$80$$ scientific calculators daily. The constraints on the manufacturing limits the production to a maximum of $$200$$ normal and $$170$$ scientific calculators. Also, a minimum of total $$200$$ calculators must be produced in a day. A normal calculator sold incurs a loss of Rs. $$2$$, whereas a scientific calculator gains a profit of Rs. $$5$$.

$$($$Take $$x$$ and $$y$$ as the quantity of normal and scientific calculators produced in a day respectively.$$)$$

...view full instructions

In order to obtain maximum profit, the quantity of normal and scientific calculators to be manufactured daily is:

A
$$(100,170)$$

B
$$(200,170)$$

C
$$(100,80)$$

D
$$(200,80)$$

Solution

let the number of normal calculators produced in a day be $$x$$ and
the number of scientific calculators produced in a day be $$y$$
the minimum of total calculators to be produced per day is $$200\implies x+y\geq 200$$

Given, the minimum number of normal calculators to be produced per day is $$100 \implies x\geq 100$$ and
the minimum number of scientific calculators to be produced per day is $$80 \implies y\geq 80$$

Also given, the maximum number of normal calculators can be produced per day is $$200 \implies x\leq 200$$ and
the maximum number of scientific calculators can be produced per day is $$170 \implies y\leq 170$$

A normal calculator incurred a loss of $$Rs.2$$
For $$x$$ normal calculators, the loss is $$Rs.2x$$

A scientific calculator gained a profit of $$Rs.5$$
For $$xy$$ scientific calculators, the gain is $$Rs.5y$$
Therefore, profit of the manufacturer $$P=5y-2x$$

substituting the options in the equation $$P=5y-2x$$ and finding the maximum value and verifying whether the option satisfies the given constraints.

substituting option A i.e., $$(x,y)=(100,170)$$
$$x+y\geq 200\implies 100+170=270\geq 200$$ True
$$x\geq 100 $$ and $$x\leq 200$$ True
$$y\geq 80 $$ and $$x\leq 170$$ True
Hence, $$P=5y-2x = 5*170-2*100=650$$

substituting option B i.e., $$(x,y)=(200,170)$$
$$x+y\geq 200\implies 200+170=370\geq 200$$ True
$$x\geq 100 $$ and $$x\leq 200$$ True
$$y\geq 80 $$ and $$x\leq 170$$ True
Hence, $$P=5y-2x = 5*170-2*200=450$$

substituting option C i.e., $$(x,y)=(100,80)$$
$$x+y\geq 200\implies 100+80=180\geq 200$$ False

substituting option D i.e., $$(x,y)=(200,80)$$
$$x+y\geq 200\implies 200+80=280\geq 200$$ True
$$x\geq 100 $$ and $$x\leq 200$$ True
$$y\geq 80 $$ and $$x\leq 170$$ True
Hence, $$P=5y-2x = 5*80-2*200=0$$

Therefore, $$(100,170)$$ satisfies all constraints and produces the maximum value than the other options.
Question 3
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The region on the graph sheet with satisfies the constraints including the non- negativity restrictions is called the _______ space.

A
solution

B
interval

C
concave

D
convex

Solution

In linear programming, the feasible region is defined as the region in which all the set of points which satisfy the given constraints and the non-negativity restrictions.

In the above figure, the blue shaded region is the feasible region which contains the solution set.
Question 4
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Directions For Questions

A company manufactures two types of calculators - normal calculator and scientific calculator. Due to demand in the market, the company must produce a minimum of $$100$$ normal and $$80$$ scientific calculators daily. The constraints on the manufacturing limits the production to a maximum of $$200$$ normal and $$170$$ scientific calculators. Also, a minimum of total $$200$$ calculators must be produced in a day. A normal calculator sold incurs a loss of Rs. $$2$$, whereas a scientific calculator gains a profit of Rs. $$5$$.

$$($$Take $$x$$ and $$y$$ as the quantity of normal and scientific calculators produced in a day respectively.$$)$$

...view full instructions

In order to maximize the profit of the company, the optimal solution of which of the following equations is required?

A
$$P = x+y-200$$

B
$$P = 5y-2x$$

C
$$P = y - 80$$

D
$$P = 200-x$$

Solution

let the number of normal calculators produced in a day be $$x$$ and
the number of scientific calculators produced in a day be $$y$$
the minimum of total calculators to be produced per day is $$200\implies x+y\geq 200$$

Given, the minimum number of normal calculators to be produced per day is $$100 \implies x\geq 100$$ and
the minimum number of scientific calculators to be produced per day is $$80 \implies y\geq 80$$

Also given, the maximum number of normal calculators can be produced per day is $$200 \implies x\leq 200$$ and
the maximum number of scientific calculators can be produced per day is $$170 \implies y\leq 170$$

A normal calculator incurred a loss of $$Rs.2$$
For $$x$$ normal calculators, the loss is $$Rs.2x$$

A scientific calculator gained a profit of $$Rs.5$$
For $$xy$$ scientific calculators, the gain is $$Rs.5y$$
Therefore, profit of the manufacturer $$P=5y-2x$$
Question 5
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Directions For Questions

A farmer has $$10$$ acres of land to plant wheat and rye. He has to plant atleast $$7$$ acres. Each acre of wheat costs $$\$200$$ and each acre of rye costs $$\$100$$ to plant. He has only $$\$1200$$ to spend. Moreover, the farmer has to get the planting done in $$12$$ hours and it takes $$1$$ hour to plant an acre of wheat and $$2$$ hours to plant an acre of rye. An acre of wheat yields a profit of $$\$500$$ and an acre of rye yields a profit of $$\$300$$.

$$($$Take $$x$$ and $$y$$ as the acres of wheat and rye planted respectively$$)$$

...view full instructions

How many acres of each (wheat and rye) should the farmer plant in order to get maximum profit?

A
$$(5,5)$$

B
$$(4,4)$$

C
$$(4,5)$$

D
$$(4,3)$$

Solution

let $$x$$ be the acres of wheat planted and
$$y$$ be the acres of rye planted

Given that there are a total of $$10$$ acres of land to plant.
Atleast $$7$$ acres is to be planted i.e., $$x+y\geq 7$$

Given that the cost to plant one acre of wheat is $$\$200$$
Therefore, the cost for $$x$$ acres of wheat is $$200x$$

Given that the cost to plant one acre of rye is $$\$100$$
Therefore, the cost for $$y$$ acres of rye is $$100y$$

Given that, amount for planting wheat and rye is $$\$1200$$
Therefore the total cost to plant wheat and rye is $$200x+100y\leq 1200\implies 2x+y\leq 12$$

Given that, the time taken to plant one acre of wheat is $$1$$ hr
Therefore, the time taken to plant $$x$$ acres of wheat is $$x$$ hrs

Given that, the time taken to plant one acre of rye is $$2$$ hrs
Therefore, the time taken to plant $$y$$ acres of rye is $$2y$$ hrs

Given that, the total time for planting is $$12$$ hrs
Therefore, the total time to plant wheat and rye is $$x+2y\leq 12$$

Given that, one acre of wheat yields a profit of $$\$500$$
Therefore, the profit from $$x$$ acres of wheat is $$500x$$

Given that, one acre of rye yields a profit of $$\$300$$
Therefore, the profit from $$y$$ acres of wheat is $$300y$$

therefore the total profit from the wheat and rye is $$P=500x+300y$$

Now substituting the options in the profit expression and verifying

Substituting option A $$(x,y)=(5,5)$$
$$x+y\geq 0\implies 5+5\geq7\implies 10\geq 7$$ True
$$2x+y\leq 12\implies 2(5)+5\leq12\implies 15\leq 12$$ False

Substituting option B $$(x,y)=(4,4)$$
$$x+y\geq 0\implies 4+4\geq7\implies 8\geq 7$$ True
$$2x+y\leq 12\implies 2(4)+4\leq12\implies 12\leq 12$$ True
$$x+2y\leq 12\implies 4+2(4)\leq12\implies 12\leq 12$$ True

Substituting option C $$(x,y)=(4,5)$$
$$x+y\geq 0\implies 4+5\geq7\implies 9\geq 7$$ True
$$2x+y\leq 12\implies 2(4)+5\leq12\implies 13\leq 12$$ False

Substituting option D $$(x,y)=(4,3)$$
$$x+y\geq 0\implies 4+3\geq7\implies 7\geq 7$$ True
$$2x+y\leq 12\implies 2(4)+3\leq12\implies 11\leq 12$$ True
$$x+2y\leq 12\implies 4+2(3)\leq12\implies 10\leq 12$$ True

$$(4,4), (4,3)$$ satisfies the constraints. Therefore finding the profit

for $$(4,4)$$, $$P=500x+300y=500(4)+300(4)=3200$$
for $$(4,3)$$, $$P=500x+300y=500(4)+300(3)=2900$$

Therefore the maximum profit is attained at $$(4,4)$$
Question 6
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Directions For Questions

A farmer has $$10$$ acres of land to plant wheat and rye. He has to plant atleast $$7$$ acres. Each acre of wheat costs $$\$200$$ and each acre of rye costs $$\$100$$ to plant. He has only $$\$1200$$ to spend. Moreover, the farmer has to get the planting done in $$12$$ hours and it takes $$1$$ hour to plant an acre of wheat and $$2$$ hours to plant an acre of rye. An acre of wheat yields a profit of $$\$500$$ and an acre of rye yields a profit of $$\$300$$.

$$($$Take $$x$$ and $$y$$ as the acres of wheat and rye planted respectively$$)$$

...view full instructions

How many acres of land will be left unplanted if the farmer aims for a maximum profit?

A
$$0$$

B
$$1$$

C
$$2$$

D
$$3$$

Solution

let $$x$$ be the acres of wheat planted and
$$y$$ be the acres of rye planted

Given that there are a total of $$10$$ acres of land to plant.
Atleast $$7$$ acres is to be planted i.e., $$x+y\geq 7$$

Given that the cost to plant one acre of wheat is $$\$200$$
Therefore, the cost for $$x$$ acres of wheat is $$200x$$

Given that the cost to plant one acre of rye is $$\$100$$
Therefore, the cost for $$y$$ acres of rye is $$100y$$

Given that, an amount for planting wheat and rye is $$\$1200$$
Therefore the total cost to plant wheat and rye is $$200x+100y\leq 1200\implies 2x+y\leq 12$$

Given that, the time taken to plant one acre of wheat is $$1$$ hr
Therefore, the time taken to plant $$x$$ acres of wheat is $$x$$ hrs

Given that, the time taken to plant one acre of rye is $$2$$ hrs
Therefore, the time taken to plant $$y$$ acres of rye is $$2y$$ hrs

Given that, the total time for planting is $$12$$ hrs
Therefore, the total time to plant wheat and rye is $$x+2y\leq 12$$

Given that, one acre of wheat yields a profit of $$\$500$$
Therefore, the profit from $$x$$ acres of wheat is $$500x$$

Given that, one acre of rye yields a profit of $$\$300$$
Therefore, the profit from $$y$$ acres of wheat is $$300y$$

therefore the total profit from the wheat and rye is $$P=500x+300y$$

In the above figure, the blue shaded region is the feasible region with three corner points.$$(4,4), (2,5), (5,2)$$

Now substituting the corner points the profit equation,
substituting $$(4,4) \implies P=500x+300y=500(4)+300(4)=3200$$

substituting $$(2,5) \implies P=500x+300y=500(2)+300(5)=2500$$

substituting $$(5,2) \implies P=500x+300y=500(5)+300(2)=3100$$

$$\$3200 $$ is the maximum profit is attained by planting $$4$$ acres of wheat and $$4$$ acres of rye.

i.e., a total of $$4+4=8$$ acres of land is planted. But, the total land is $$10$$ acres. therefore, the unplanted land is $$10-8=2$$ acres
Question 7
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If $$a,b,c \in +R$$ such that $$\lambda abc$$ is the minimum value of $$a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)$$, then $$\lambda=$$

A
$$1$$

B
$$3$$

C
$$4$$

D
None of the above.

Solution

We know that $$A.M.\geq G.M.$$

Therefore, $$\dfrac{b^2+c^2}{2}\geq \sqrt{b^2c^2}$$

$$\implies b^2+c^2\geq 2bc$$

Multiplying $$a$$ on both sides doesn’t change the inequality. Since, given that $$a$$ is positive.

$$\implies a(b^2+c^2)\geq 2abc$$ ————(1)

Similarly, $$b(a^2+c^2)\geq 2abc$$ ————(2)

and $$c(a^2+b^2)\geq 2abc$$ —————(3)

adding (1), (2) and (3) we get

$$a(b^2+c^2)+b(a^2+c^2)+c(a^2+b^2)\geq 2abc+2abc+2abc$$

$$\implies a(b^2+c^2)+b(a^2+b^2)+c(a^2+b^2)\geq 6abc$$

Therefore $$\lambda$$ is $$6$$
Question 8
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Consider the objective function $$Z = 40x + 50y$$ The minimum number of constraints that are required to maximize $$Z$$ are

A
$$4$$

B
$$2$$

C
$$3$$

D
$$1$$

Solution

Since in the given function $$Z=40x+50y$$, two variables are used.

So, the two constraints will be $$x \geq 0, y \geq 0$$ and the third one will be of the type $$ax + by \geq c$$.

Hence, at least $$3$$ constraints are required.
Question 9
1 / -0

Maximum value of $$z=3x+4y$$ subject to $$x-y\le -1,-x+y\le 0,x,y\ge 0$$ is given by?

A
$$1$$

B
$$4$$

C
$$6$$

D
None of the these

Solution

No common feasible area
No commonpoint,
So, $${ Z }_{ max }$$ cant find for these curves.
Hence none of these.
Question 10
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The objective function $$z={ x }_{ 1 }+{ x }_{ 2 }$$, subject to $${ x }_{ 1 }+{ x }_{ 2 }\le 10,{ -2x }_{ 1 }+3{ x }_{ 2 }\le 15,{ x }_{ 1 }\le 6$$, $${ x }_{ 1 },{ x }_{ 2 }\ge 0$$ has maximum value .................. of the feasible region.

A
at only one point

B
at only two points

C
at every point of the segment joining two points

D
at every point of the line joining two points

Solution

For the equation, $$x_1+x_2=10$$,
$$x$$ $$0$$ $$10$$ $$5$$
$$y$$ $$10$$ $$0$$ $$5$$
For the equation, $$-2x_1+3x_2=15$$,
$$x$$ $$0$$ $$-3$$ $$3$$
$$y$$ $$5$$ $$3$$ $$7$$
The feasible region for the given lines is as indicated in the figure.
The values of $$z$$ at the corner points are:
At $$(0,0)$$, $$z = 0$$
At $$(0,5)$$, $$z = 5$$
At $$(3,7)$$, $$z = 3+7 = 10$$
At $$(6,4)$$, $$z = 6+4 = 10$$
At $$(6,0)$$, $$z = 6$$
Hence, the objective function $$z$$, has maximum value at every point of the segment joining two points of the feasible region.

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$$x$$	$$0$$	$$10$$	$$5$$
$$y$$	$$10$$	$$0$$	$$5$$

$$x$$	$$0$$	$$-3$$	$$3$$
$$y$$	$$5$$	$$3$$	$$7$$

Linear Programming Test - 24

Result

Linear Programming Test - 24

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Rank

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SHARING IS CARING

Question 1

origin

corner points

centre

edge

Solution

Question 2

$$(100,170)$$

$$(200,170)$$

$$(100,80)$$

$$(200,80)$$

Solution

Question 3

solution

interval

concave

convex

Solution

Question 4

$$P = x+y-200$$

$$P = 5y-2x$$

$$P = y - 80$$

$$P = 200-x$$

Solution

Question 5

$$(5,5)$$

$$(4,4)$$

$$(4,5)$$

$$(4,3)$$

Solution

Question 6

$$0$$

$$1$$

$$2$$

$$3$$

Solution

Question 7

$$1$$

$$3$$

$$4$$

None of the above.

Solution

Question 8

$$4$$

$$2$$

$$3$$

$$1$$

Solution

Question 9

$$1$$

$$4$$

$$6$$

None of the these

Solution

Question 10

at only one point

at only two points

at every point of the segment joining two points

at every point of the line joining two points

Solution

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