x +3 y=6 andx+4 y=4, weget x= 12

5 , y= 2 5

Hence vertex is (12/5,2/5)

306 (d)

Common region is quadrilateral.

307 (d)

We draw the lines

x+2 y =8 by joining two points A(8,0) and B(0,4)

2 x +2 y=9

by joining two points C(9/2,0) and D(0,9/2) 2 x + y =7 by joining two points E(7/2,0) and F(0,7/2)

The point P point of intersection of the lines

2 x +2 y=9∧2 x+ y =7

solving these equation

x=5/2, y=2 P=(5/2,2)

The point Q is the point of intersection of the lines x+2 y =8∧2 x +2 y =9 solving these equation we get

x=1, y=7/2

Q=(1,7/2)

The point Q(1,7/2) =(1,3.5) provide the solution to the L.P.P.

308 (c)

We draw the lines

5 x+ y =10 by joining two points A(2,0) and B(0,10)

2 x +2 y=12

by joining two points C(6,0) and D(0,6) x+4 y =12 by joining two points E(12,0) and F(0,3)

The point G point of intersection of the lines

x+4 y =12∧2 x+2 y=12

solving these equation

x=4, y=2

G=(4,2)

The point H is the point of intersection of the lines

2 x +2 y=12∧5 x + y=10

solving these equation we get

x=1, y=5 H=(1,5)

The shaded region EGHB is a feasible region Optimal solution is obtained at one of the points E(12,0), G(4,2), H(1,5), B(0,10)

Corner Points value of

z=4 x+5 y

E(12,0) p=48+0=48

G(4,2) p=16+10=26 H(1,5) z=4+25=29 B(0,10) z=0+50=50

The minimum value of p is 48 at the point G(4,2)

309 (d)

We draw the lines

2 x + y =9 by joining two points A(9/2,0) and B(0,9/2)

x+2 y =9

by joining two points C(9,0) and D(0,9/2) x+ y=7 by joining two points E(7,0) and F(0,7)

P is the point of intersection of the lines

x+2 y =9 andx+ y=7

solving these equation x=5, y=2

G=(5,2)

The point Q is the point of intersection of the lines

x+ y=7∧2 x + y=9

solving these equation we get

x=2, y=5 Q=(2,5)

The shaded region CPQB is a common region

∴

The corner points are C(9,0), P(5,2), Q(2,5), B(0,9)

310 (c)

Option :(a)

Shaded region represents given inequality and is a convex set Option :(b)

Shaded region represents given inequality and is a convex set Option :(c)

x=5 → x=± 5

∴

points on the lines

x=5, x =−5

This is not a convex set.

Option :(d)

3 x

²

+2 y

_❑²

≤ 6

_i.e.

x

²

/2+ y

_❑²

/3 ≤ 1

Shaded region represents given inequality and is a convex set

329 (c)

Z =3 x +2 y

Zat (0,2)=3 (0)+2 (2)=4 Zat (0,0)=3 (0)+2 (0)=0

0=¿

Zat

(

_2,0

)

₌₃

(

₂

)

_+2¿

∴ Zismaximizeat(2,0)

∴( c) iscorrectanswer .

330 (b)

Z =3 x +2 y

Zat (0,3)=3 (0)+2 (3)=6 Zat (0,1)=3 (0)+2 (1)=2 Zat ( ² 3 , 7

3 ) ⁼³ ( 3 ² ) ⁺² ( ⁷ 3 ) ^{= 20} 3

∴ Zismaximizeat( 2 3 , 7

3 )

∴( B) iscorrectanswer .

331 (d)

Z =9 x+13 y

Zat (0,6)=9 (0 )+ 13(6 )=78 Zat (5,0 )=9 (5)+13 (0)=45 Zat (3,4 )=9(3 )+13 ( 4)=79

∴ Zismaximizeat (3,4 )

∴( D) iscorrectanswer .

332 (b)

P=30 x +20 y

Pat (0,3)=30 (0)+20 (3)=60 Pat (1,3/2)=30 (1)+20 (3 /2)=60

Pat (4,0)=30( 4)+20 (0 )=120 Pat (0,8)=30 (0 )+20 (8 )=160 Pat (8,0)=30 (8 )+20 (0 )=240

∴ Zisminimiumat (1,3/2)

∴( B) iscorrectanswer .

339 (a)

Function

x

²

+2 x +3

satisfies all the points.

340 (a)

200 x+100 y=2000

i. e .2 x+ y=20 …(1)

Represent a line through the points A(10,0) and B(0,20) x+2 y =50 …(2)

Represent a line through the points C(50,0) and D(0,25)

40 x +40 y=1400

i. e . x + y=35 … (3)

Represent a line through the points E(35,0) and F(0,35) Lines (2) and (3) intersect at points G(20,15)

Feasible region is the shaded portion FGC with vertices F(0,35),G(20,15),C(50,0)

z=4 x+8 y

At F(0,35), Z=4(0)+8(35)=280 At G(20,15), Z=4(20)+8(15)=200 At C(50,0), Z=4(50)+8(0)=200 At D(3,0),=P=60

At A(0,0),=P=0

∴ minimum value of Z is 200 at seg GC

341 (b)

Consider the equation, 2 x − y=1

We may write it as

x 1 2

+ y

−1 =1

This shows the line

2 x − y=1

makes intercepts of ½ and -1

On the axes. Thus, the line meets X-axis at (1/2,0) and Y-axis at (0,-1). We plot these points and join them by thick line.

Consider (0, 0).Clearly (0, 0) does not satisfy the given in equation.

Therefore, out of portions divided by this line, the one not containing (0, 0) together with the points on the line forms a solution set.

342 (b)

Consider the equation,

x+ y=1

It meets the axes at A (1, 0) and B (0, 1).

Join these points.

Clearly, the portion not containing (0,0) is the solution set of x+ y ≥ 1 Consider

7 x+9 y=63 i. e . x

9 + y 7 =1

Thus, the line meets axes at C (9,0) and D(0,7).

Join these points.

Clearly the portion containing (0,0) is the solution set of

7 x+9 y ≤ 63

Y=5, is a line parallel to X-axis at the distance 5 from X-axis and the portion containing (0,0) is the solution set of the in equation

y ≤ 5.

x=6, is a line parallel to Y-axis at the distance 6 from Y-axis and the portion containing (0,0) is the solution set of the in equation

x ≤ 6.

Clearly, x ≥ 0 has a solution represented by Y-axis and the portion on its right.

Also,

y ≥ 0

a solution represented by X-axis and the portion above it.

The shaded region represents the solution of the given system of inequations.

343 (b)

X:distance travelled with speed 25km/hr y:distance travelled with speed 40km/hr

∴ maximizez=x+ ysubjecttoconstraints

2 x +5 y ≤100, x

25 + y

40 ≤1, x ≥ 0, y ≥ 0 2 x +5 y=100 … (1)

Represents a line through the points A(50,0) and B(0,20)

x

25 + y

40 =1 … (2)

Represents a line through the point C(25,0) and D(0,40) Lines (1) and (2) interacts the point E(50/3,40/3)

The feasible region is OCEB with corner point O(0,0),C(25,0),E(50/3,40/3),B(0,20) At O(0,0) Z=0+0=0

At C(25,0) Z=25+0=25 At E(50/3, 40/3) Z=50/3+40/3=30 At B(0,20) Z=0+20=20

∴ Maximum distance covered in 1 hour is 30 km with 50/3 km at 25km/hr and 40/3 km at 40 km/hr.

344 (b)

Let x kg of wheat and y kg of rice be mixed.

Then we have to minimize the cost function C=2 x +8 y subject to the constraints given below.

X kg of wheat and y kg of rice must give at least 50 gm i.e.0.05 kg of proteins.

∴ 0.1 x+0.05 y≥ 0.05

∴10 x+5 y≥ 5∨2 x + y≥ 1

Similarly, x kg of wheat and y kg of rice must give at least 200gm i.e.0.2kg of carbohydrates.

∴ 0.25 x+0.5 y ≥0.2

∴25 x +50 y≥ 20∨5 x +10 y ≥ 4

Thus, we have to minimize

C=2 x +8 y

subject to the constraints

x ≥ 0, y ≥ 0, 2 x + y ≥1, 5 x+10 y ≥ 4

meets the axis at A(1/2,0) and B(0,1)

5 x+10 y=4 … (2)

meets the axis at C(4/5,0) and D(0,2/5)

x=0 is the y-axis and y=0 is the x-axis.

Lines(1) and (2) meets at E (6/15,3/15)

The feasible region is DEB with corner point D(0,2/5),E(6/15,3/15),B(0,1) The vales of c=2 x+8 y at these points are 3.2, 2.4 and 8 respectively.

The minimum value of

c=2 x+8 y

is 2.4 obtained when x=2/5 and y=1/5 i.e. when x=400 gm and y=200 gm .then the diet cost is minimum.

345 (c)

X=2 represents a line through the points A (2, 0) and parallel to Y-axis.

y=3 represents a line through the points B (0, 3) and parallel to X-axis.

Line

x+ y=4

meets the axes at C(4,0) and D(0,4)

x=2 andx+ y=4 Intersect at the point E (2,2) and the lines y=3 andx + y=4 intersect at the point F(1,3).

∴

The shaded portion OAEFB is the feasible region with vertices O(0,0),A(2,0),E(2,2),F(1.3),B(0,3).

P=3 x +5 y

P(A)=3(2)+5(0)=6 P(E)=3(2)+5(2)=16 P(A)=3(1)+5(3)=18 P(A)=3(0)+5(3)=15

Hence the maximum value of p is 18.

346 (d)

2 x + y =4 meets the axes at A(2,0) and B(0,4)

x+2 y =4

meets the axes at C(4,0) and D(0,2) Lines 2 x + y =4 andx+2 y=4 intersect at the points E(4/3,4/3)

∴

The feasible region is shaded portion OAED with vertices O(0,0),A(2,0),E(4/3,4/3),D(0,2)

P= x+y P(A)=2+0=2

P(E)=(4/3+4/3)=(8/3) P(D)=0+2=2

∴ Maximum value of P is (8/3).

347 (a)

2 x + y =6

meets the axes at A(3,0) and B(0,6) x+ y=4 meets the axes at C(4,0) and D(0,4) X=y represents a line through origin.

Hence only those points which lie in the region between 2x+y=6 and x=y are feasible.

Lines 2x+6=6 and x=y intersect at the points E (2, 2) which is obvious that the point at which E (2, 2) =2+3(2) =8 is least.

348 (b)

x+ y=4

meets the axes at A(4,0) and B(0,4) 3 x+8 y =24 meets the axes at C(8,0) and D(0,3)

10 x+7 y =35

meets the axes at E(35/10,0) and F(0,5) Lines 10x+7y=35 and x+y=4 intersect at point G(7/3,5/3) Lines 3x+8y=24 and x+y=4 intersect at point H(8/5,12/5)

∴ The shaded portion OEGHD is the feasible region with vertices O(0,0), E(3.5,0),G(2.4,1.6), H(1.6,2.4), D(0,3)

The value of z=5x+5y at these points are Z(E)=17.5

Z(G)=20 Z(H)=20 Z(D)=15

∴

The maximum value is 20.

349 (b)

x+2 y =2000

meets the axis at A(2000,0) and B(0,1000) x+ y=1500 meets the axis at C(1500,0) and D(0,1500)

X=600 represents a line parallel to Y-axis passing through E(600,0) Lines x=600 and x+2y=2000 intersect at point F(600,700)

∴

The shaded portion OEFB is the feasible region with vertices O(0,0), E(600,0),F(600,700), B(0,1000), D(0,3)

The corresponding value of z=3x+5y at these points are Z(E)=1800

Z(G)=5300 Z(H)=5000

∴ The maximum value is 5300.

350 (d)

4 x +2 y=46 meets the axis at the points A(23/2,0) and B(0,23) X+3y=24 meets the axis at the points C(24,0) and D(0,8)

Lines 4x+2y=46 and x+3y=24 intersect at point E(9,5)

∴

The shaded portion OAED is the feasible region with vertices A(23/2,0),E(9,5), D(0,8)

The corresponding values of P=4x+2y are P(A)=46

P(E)=46 P(D)=16

∴ Hence The maximum value is 46.

351 (c)

x+ y=8

meets the axis at A(8,0) and B(0,4) 6 x+8 y=12 meets the axis at C(2,0) and D(0,3)

5 x+8 y =20

meets the axis at E(4,0) and F(0,20/8)

Lines 6x+4y=12 and 5x+8y=20 intersect at point P(4/7,15/7)

∴ The shaded portion EABGDP is the feasible region with vertices E(4,0), A(8,0),B(0,8), D(0,3), P(4/7,15/7)

The value of function z=30x+20y at the above vertices are Z(E)=120

Z(A)=240 Z(B)=160 Z(D)=60 Z(P)=60

∴

The Z has minimum value 60 at P(4/7,15/7) and C(0,3).

352 (a)

x+y=5 …(1)

Represent a line through the points A(5,0) and B(0,5) x+2y=4 …(2)

Represent a line through the points C(4,0) and D(0,2) 4x+y=12 …(2)

Represent a line through the points E(3,0) and F(0,12) Lines (2) and (3) intersect a point G(20/7,4/7)

Lines (1) and (3) intersect a point H(7/3,8/3)

The feasible region is the shaded portion DGHB with vertices D(0,2),G(20/7,4/7),H(7/3,8/3),B(0,5)

Values of objective function z=6x+3y at these vertices are

z ( D)=6 ×0+3 × 2=6

Z ( D)=6 × 0+3 × 4 7 = 132

7 Z ( D)=6 × 7

3 +3 × 8 3 =22 z ( B)=6 × 0+3 ×5=15

∴

maximum value of Z is 22 at (7/3,8/3)

353 (b)

x=3 Represents a line through the points A (3, 0) and parallel to Y-axis.

y=3 Represents a line through the points B (0, 3) and parallel to X-axis.

x+y=5 meets the axes at point C(5,0) and D(0,5) 2x+y=4 meets the axes at point E(2,0) and F(0,4) Lines x=3 and x+y=5 intersect at point G(3,2) Lines y=3 and x+y=5 intersect at point H(2,3) Lines y=3 and 2x+y=4 intersect at point I(1/2,3)

The shade portion EAGHI is feasible region with vertices E(2,0),A(3,0),G(3,2),H(2,3) and I(1/2,3)

Values of objective function z=3x+2y at these vertices are Z(E)=6

Z(A)=9 Z(G)=13 Z(H)=12 Z(I)=15/2

∴

Z has maximum value 13 at C(3,2)

354 (c)

2x+3y=6 meets the axes at points A(3,0) and B(2,0) 2x+y=4 meets the axes at point C(2,0) and D(0,4)

x=4 Represents a line through the points E (4, 0) and parallel to Y-axis.

y=6 Represents a line through the points F (0, 6) and parallel to X-axis.

Lines y=6 and x=4 intersect at point G(4,6)

Lines 2x+y=4 and 2x+3y=6 intersect at point H(3/2,1)

The shade portion AEGFDH is feasible region with vertices A(3,0), E(4,0),G(4,6),F(0,6),D(0,4) and H(3/2,1)

Values of objective function z=3x+2y at the above vertices are

Z(A)=9 Z(B)=12 Z(G)=24 Z(F)=12 Z(D)=8 Z(H)=13/2

∴

Z has minimum value 13/2

355 (c)

2 x

₁

+x

₂

=104

meets the axes at points A(52,0) and B(0,104)

x

₁

+2 x

₂

=76

meets the axes at points C(76,0) and B(0,38)

The origin (0, 0) satisfy both the given in equalities, so the regions below the two lines containing the origin are the regions satisfy these constants.

Lines

2 x

₁

+ x

₂

=104∧x

₁

+ 2 x

₂

=76

intersect at point E(44,16)

The shaded portion OAED is feasible region with vertices O(0,0),A(52,0), E(44,16), D(0,38)

Values of objective function

z=6 x

₁

+11 x

₂ at the above vertices are Z(O)=0

Z(A)=312 Z(E)=440 Z(D)=418

∴ The maximum value of z is 440.

356 (c)

In the given figure origin is present in shaded area, so equations are 2 x +5 y ≤80, x+ y ≤ 20, x ≥ 0, y ≥ 0

357 (a)

The line passes through origin and it represents the half plane that contains the positive X-axis.

358 (b)

Required region is rectangle ABCD

Point A is point of intersection of -3x+2y=3 and 6x+4y=24

∴ A ≡( 3 2 , 15

14 )

Point B is point of intersection of -3x+2y=3 and 2x+3y=6

∴ B ≡( 3 13 , 24

13 )

Point C is point of intersection of x-2y=2 and 2x+3y=6

∴C ≡( 18 7 , 2

7 )

Point D is point of intersection of x-2y=2 and 6x+4y=24

∴ D ≡(7 2 , 3

4 )

Z=5x+2y

atA ≡ ( ³ 2 , 15

14 ) , z=15, AtB ≡ ( 13 ³ , 24

13 ) ^{, z=4.84}

atC ≡ ( ¹⁸ 7 , 2

7 ) , z=13.42, AtD ≡ ( ⁷ 2 , 3

4 ) ^{, z =19}

∴ for the maximum value of Z=5x+2y, x=7/2, y=3/4

359 (a)

Let x unit of food A and y unit of food B be used to give the sick person the least quantities of vitamins, mineral and calories.

Total cost of food is C=4x+3y which is to be minimized.

For x unit of food A and y units of food B a total of 200x+100y units of vitamins is obtained and its minimum requirement is 1400.

∴ we have200 x+100 y ≥1400 Similarly for minerals, we have:

x+2 y ≥50

And for calories 40 x +40 y ≥1400

Since the number of units purchases cannot be negative.

x ≥ 0, y ≥ 0

After plotting the constraints we get the feasible region as FGC with vertices F(0,35), G(20,15) and C(50,0)

Values of C=4x+3y at above vertices are C(F)=4(0)+3(35)=105

C(G)=4(20)+3(15)=125 C(H)=4(50)+3(0)=200

Thus the minimum cost is 105 at x=0 and y=35.

360 (c)

Let A=number of product A and y=number of product B, which the carpenter produces.

Total profit is p=30x+20y which is to be maximized subject to the constraints

2 x + y ≤20, x +3 y ≤ 15, x ≥ 0, y ≥ 0

Now 2x+y=20 …(1)

i. e . x 10 + y

2 =1

is a line through the points A(10,0) and B(0,20) Now x+3y=15 …(1)

i. e . x 15 + y

5 =1

is a line through the points C(15,0) and D(0,5) Moreover lines (1) and (2) intersect at R(9,2)

Clearly shaded the portion OARD in the figure is the feasible and its corner points are O(0,0), A(10,0), R(9,2) and D(0,5)

At O(0,0) P=30(0)+20(0)=0 At A(10,0) P=30(10)+20(0)=300 At R(9,2) P=30(9)+20(2)=310 At O(0,0) P=30(0)+20(5)=100

∴ maximum profit is Rs.310 at R(9,2)

i.e. when 9 units of product A and 2 units of product B are produced.

361 (a)

x+4y=4 …(1)

Represents a line through the points A(4,0) and B(0,1) 2x+3y=6 …(1)

Represents a line through the points C(3,0) and D(0,2) Lines (1) and (2) intersect at the points P(12/5,2/5) Also, x ≥ 0, y ≥ 0.

The feasible region of the given L.P.P. is the shaded region BPD of figure and its corner points are B(0,1), P(12/5,2/5), D(0,2)

The objective function is f(x,y)=x+2y F(0,1)=0+2=2

f(0,2)=0+2(2)=4

Clearly maximum value of f(x,y) is 4 at (0,2) and its minimum value is 2 at (0,1).

362 (d)

Draw the lines x+y=4, 3x+4y=12 Shaded region is feasible region

The vertices are A (4,0), D(0,3), O(0,0).

The value of objective function Z=5x+7y at the vertices.

Z(O)=0

Z(A)=5x0+7x3=21

∴

Z has maximum value 21 at (0,3)

363 (b)

x+2y=4 …(1)

Represents a line through the points A(4,0) and B(0,2) x-y=-2 …(2)

Represents a line through the points C(-2,0) and B(0,2) Lines (1) and (2) intersect at the points B(0,2)

x=2 …(3)

Is a line parallel to Y-axis through point D(2,0) and it meets lines (1) and (2) in the points P(2,1) and Q(2,4) respectively.

y=-2 …(4)

Is a line parallel to X-axis through point E(0,-2) and it meets lines (1) , (2) and (3) in the points L(8,-2) M(-4,-2) and N(2,-2) respectively.

The feasible region of the given L.P.P. is the convex polygon MNPB(shaded portion) in the figure and its corner points are M(-4,-2), N(2,-2), P(2,1),B(0,2)

The objective function is f(x,y)=x+2y f(-4,-2)=-4-4=-8

f(2,-2)=2-4=-2 f(2,1)=2+2=4 f(0,2)=0+4=4

Clearly maximum value of ‘f’ is 4 at (2, 1) and (0,2) and minimum value f is -8 at (-4,-2).

Let

x

₁

, x

₂ denote the number of questions to be attempted from section A, B respectively.

Then mathematically the problem can be written as maximize

M=10 x

₁

+15 x

_{2 [Total} marks] subject to constraints;

x

₁

≥ 2, x

₂

≥ 3, 15

_x1

+ 25 x

₂

≤ 180,

[3 hours = 180 minutes]

x

₁

+x

₂

≤10∧x

₁

≥ 0, x

₂

≥ 0

Consider a set of rectangular Cartesian axes OX1, OX2 in the plane.

Each point has coordinate of the type

( x

₁

, x

₂

)

and conversely. Clearly any point

( x

₁

, x

₂

)

which satisfies the conditions

x

₁

≥ 0, x

₂

≥0,

lies in the first quadrant only. Thus our search for the desired number

pair

x (¿¿ 1 , x

2

)

¿

Will be restricted to the first quadrant only. Also since

x

₁

≥ 2, x

₂

≥ 3,

15

_x1

+ 25 x

₂

≤ 180, x

₁

+ x

₂

≤10

∴ The desired point

( x

₁

, x

₂

)

lies some where in the convex region GHI which is shown (shaded) in the figure. This is feasible region i.e. solution space of the problem.

The extreme points are G(2,3), H(7,3), I(2,6) At G(2,3), M=10(2)+15(3)=65

At H(7,3), M=10(7)+15(3)=115 At I(2,6), M=10(2)+15(6)=110

Since the maximum value occurs at H,

∴ x

₁

=7

is the optimal solution.

x

₂

=3∧115

is the maximum marks.

365 (a)

2 x

₁

+7 x

₂

=22

meets the axes at points A(11,0) and B(0,22/7)

x

₁

+x

₂

=6

meets the axes at points C(6,0) and D(0,6)

5 x

₁

+ x

₂

=10

meets the axes at points E(2,0) and F(0,10) Lines

2 x

₁

+ 7 x

₂

=22∧x

₁

+ x

₂

=6

intersect at point G(4,2) Lines

5 x

₁

+ x

₂

=10∧x

₁

+ x

₂

=6

intersect at point H(1,5)

The shaded region FHGA is feasible region with vertices F(0,10),H(1,5), G(4,2), A(11,0)

Values of objective function

z=2 x

₁

+ 3 x

₂ at the above vertices are Z(F)=30

Z(H)=17 Z(G)=14 Z(A)=11

∴

The minimum value is 14.

366 (c)

Let process I and II be x and y times. Then 5 x+6 y ≤ 250,3 x +5 y ≤ 200

5 x+4 y ≤150, 8 x+4 y ≥ 80, x ≥0, y ≥0

Clearly 8 x+4 y ≥ 80 does not affect.

So, the number of constraints is 5.

367 (b)

2 x + y =2i . e . x 1 + y

2 =1

This line meets the axes at (1,0) and (0,2).

x− y=3 i . e . x 3 + y

−3 =1

This line meets the axes at (3,0) and (0,-3) To find common vertex solve

2x+y=2 …(i) And x-y=3 …(iI)

On solving equations (i) and (ii), we get

x= 5

3 , y= −4 3

Thus common vertex of linear in equalities is

( 5 3 ,− 4

3 )

368 (b)

The time taken in two simple question ¿

2× 3

2 = 3

minute and in one difficult questions =3 minute.

∴ For three successive questions total time taken= 6 minute.

→ Total solved question in

3 hours= 180

6 ×3=30 ×3=90

_.

369 (a)

i=1, x

₁₁

+ x

₁₂

+ x

₁₃

+...+x

_{1 n}

i=2, x

₂₁

+ x

₂₂

+x

₂₃

+...+ x

_{2 n}

i=3, x

₃₁

+ x

₃₂

+ x

₃₃

+ ...+x

_{3 n}

……….

i=m , x

_{m 1}

+ x

_{m 2}

+ x

_{m 3}

+...+x

_mn

Thus number of constraints is m (II) condition

j=1, x

₁₁

+x

₂₁

+ x

₃₁

+...+ x

_{m 1}

j=2, x

₁₂

+ x

₂₂

+ x

₃₂

+...+x

_{m 2}

……….

j=n , x

_{1 n}

+ x

_2n

+ x

_{3 n}

+...+x

_mn

Thus number of constraints is n

∴

Total number of constraints = m+n

370 (d)

In the given figure the feasible region for given constraints is line segment GI.

371 (d)

A column in the simplex table that contains all the variables in the solution is called pivot or key column.

372 (c)

(A) (B)

(C) (D)

From the above figure only C is not a convex polygon.

380 (b)

Consider the equation,

2 x − y=1

On the axes. Thus, the line meets X-axis at (1/2,0) and Y-axis at (0,-1). We plot these points and join them by thick line.

Consider (0, 0).Clearly (0, 0) does not satisfy the given in equation.

Therefore, out of portions divided by this line, the one not containing (0, 0) together with the points on the line forms a solution set.

381 (c)

x=2 represents a line through the points A (2, 0) and parallel to Y-axis.

y=3 represents a line through the points B (0, 3) and parallel to X-axis.

Line

x+ y=4

meets the axes at C(4,0) and D(0,4)

x=2 andx+ y=4 Intersect at the point E (2, 2) and the lines y=3 andx + y=4 intersect at the point F (1, 3).

∴

The shaded portion OAEFB is the feasible region with vertices O(0,0),A(2,0),E(2,2),F(1,3),B(0,3).

Hence the maximum value of p is 18.

382 (a)

If we puts the given points in the given constraints, we find that (2,3), (3,1), (2,2) are not satisfying x+2 y ≥11 _because

¿

( 2,3) ,(2+2 ×3 )≥ 11isfalse

¿

( 3,1), (3+3 ×1 )≥ 11isfalse

¿

( 2,2) ,(2+2 ×2) ≥11isfalse

¿

( 3,4) , issatisfyingas 3+2× 4 ≥ 11istrue

383 (c)

We draw the graph of

x

₁

≤ 4, x

₂

≤ 6, x

₁

6 + x

₂

9 ≤ 1

Different point of convex polygon are (0,0), (4,0), (4,3), (2,6) and (0,6) Maximum value of

z=3 x

₁

+5 x

₂ is at (2,6),z=3x2+5x6=36

∴ x

₁

=2, x

₂

=6, z=36 isthesolution.

384 (a)

A set S is called convex if for all

x

₁

, x

₂

, Sand ϵ 0 ≤ λ ≤1, λ x

₁

+ (1−λ) x

₂

ϵ . S

The set

{x : x =5 }

is not a convex set, can be seen by drawing the region.

386 (d)

On drawing graphs of given in equations, We have

Now, Z=3x+4y

Z

_(0,0)

= 0

Z

_(40,0)

=120++0=120 Z

_(0,30)

=0++120=120 Z

_(20,20)

=60++80=140

∴

Hence the maximum value =140.

Which is obtained at x=20, and y=20

387 (c)

Graph of given constraints is shown below:

Co-ordinates of B from 2x+2y=9 and 2x+y=7 are (2.5,2)

Co-ordinates of C from 2x+2y=9 and x+2y=8 are (1, 3.5).

value of objective function

P=2x+3y at various points P(0,0)=0+0=0, P(2.5,2)=11, P(3.5,0)=7, P(1,3.5)=12.5

∴

P (max.)=12.5 which is at (1,3.5)

388 (d)

The L.P.P. in standard form is Maximize

z=21 x

₁

+15 x

₂

+ 0 s

₁

+0 s

₂

Subject to ,

x

₁

+2 x

₂

+ s

₁

=6, 4 x

₁

+3 x

₂

+ s

₂

=12, x

₁

, x

₂

≥ 0, s

₁

, s

₂

≥0

x

₁

∧x

₂ are decision variable.

s

₁

∧ s

₂ are slack variables are n=4 and number of equations are m=2.

∴ The total possible basic solution are nCm=4 C 2=6

389 (c)

Obviously the ans is (C)

391 (b)

For maximization L.P. model, the simplex method is terminated when all values

c

_j

−z

_j are less than or equal to zero.

392 (a)

(3,1), (2,0) are vertices of Min z for (2,0)

Hence

x

₁

=2

393 (a)

c

394 (c)

The graph of linear programming problem is as given below.

Hence the required feasible region by the graph whose vertices are A(1.2,2.6), B(4.5,1.5) and C(8/3,10/3)

Thus objective function is minimum at A(1.2,2.6) so

x

₁

=1.2, x

₂

=2.6 andz =2× 1.2+3× 2.6=10.2

395 (d)

Clearly point D is outside.

396 (b)

A= ( ^{0, 134} 3 ) ^{, B (}

^¿

0,40) andC (10,38)

Max z for C i.e. =40+342=382.

397 (d)

The shaded region represents the bounded region.

(3, 3) satisfies, so x=3, y=3 and z=15.

399 (a)

The equations, corresponding to in equalities

3 x+2 y ≤6∧6 x +4 y ≥ 20 are 3 x +2 y=6∧6 x +4 y =20.

So the line represented by these equations are parallel.

400 (c)

Solving 2x+3y=6 and x+4y=4, we get

x= 12

5 , y= 2 5

Hence vertex is

( 12 5 , 2

5 )

402 (d)

Min z= 2(0)+ 2 (40)80

403 (a)

Min z= 2(2)+2(3)=C=10

404 (b)

To test the origin for 2x+y=2, x-y=1, and x+2y=8 in reference to shaded area,

0+0<2

is true for 2x+y=8 in reference to shaded area,

0+0<2

is true for 2x+y=2

So for the region does not include origin (0,0), 2 x + y ≥ 2 Again for x-y=1,

0−0<1

∴ x− y≤ 1

Similarly for x+2y=8,

0+0<8

∴ x +2 y ≤ 8

405 (d)

It is obvious

406 (a)

It is obvious

407 (b)

For Max. profit z=40x+25y

408 (c)

6 x+10 y ≥ 60, 4 x+3 y ≥ 40

As min expenditure is concerned.

409 (a)

15x+20y

410 (a)

x+ y ≤

(

80 ×60=480

)

∧2 x + y ≤(10 ×60=600)

411 (a)

The linear constraints are

x+2 y ≤ 400, x + y ≤300 andx , y ≥ 0

The graph for above questions.

Hence the vertices of feasible region are (300,0), (0,200), (200,100)

413 (b)

x ≥ 0, y ≥ 0, 3 x +5 y ≤80, 3 x+3 y ≤ 50

414 (c)

x+ y ≤ 100 ;4 x +9 y ≤300 ;100 x+120 y =c

415 (c)

x+ y ≤ 5,2 y−x ≤ 4, x ≥ 0, y ≥ 0

416 (a)

z=(x+2y)

Min z=0+4(2)=8

418 (b)

2 x + y ≤30, x +2 y ≤ 24, andx , y ≥ 0 is ,

The shaded region represents the feasible region hence P=6x+8y.obviously it is maximum at (12,6)

Hence P=12 ×6+8× 6=120

419 (c)

Obviously, P=x+3y will be maximum at (0,10)

∴ P=0+3 ×10=30

420 (c)

Obviously, the optimal solution is found on the line which is parallel to ‘isoprofit line’.

Hence it has infinite number of solutions.

421 (b)

Obviously, max. 4x+5y=95. It is at (5,15)

424 (b)

The given inequality is

2 x + y >5

∴2 x+ y−5>0

Substituting origin in above equation, we get

2× 0+0−5>0

∴−5>0

It does not satisfied the inequality Inequality does not contain origin

Required solution set is open half plane not containing the origin.

425 (c)

Constraints are

2 x +3 y ≤36, 5 x+2 y ≤50, 2 x +6 y ≤ 60, x ≥ 0, y ≥0

∴Thenumberofconstraintsare 5.

426 (a)

The inequality is

3 x+4 y ≤12

∴ x 4 + y 3 ≤ 1

∴ The half plane containing the origin and the points of the line 3 x+ 4 y=12 _is the required solution set.

427 (a)

The feasible region is the shaded portion with vertices (1,0), (4,0), (4,4), (3/4,3/4)

In document 10) Linear Programming Problems (Page 136-170)