Gujarat Board GSEB Textbook Solutions Class 12 Maths Chapter 12 Linear Programming Ex 12.1 Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 12 Maths Chapter 12 Linear Programming Ex 12.1

Solve the following Linear Programming Problems graphically:

Question 1.

Maximise Z = 3x + 4y

subject to constraints: x + y ≤ 4, x ≥ 0, y ≥ 0.

Solution:

Objective function is

Z = 3x + 4y

Constraints are

x + y ≤ 4, x ≥ 0, y ≥ 0

(i) consider the line

x + y = 4.

It passes through the points (4, 0) and (0, 4).

Putting x = 0, y = 0 in x + y ≤ 4, we get 0 ≤ 4 which is true.

Therefore, origin lies in the region x + y ≤ 4.

x ≥ 0 is the region to the right of y-axis and y ≥ 0 is the region on and above x-axis.

∴ The feasible region is ∆ OAB.

At (4, 0), Z = 3x + 4y = 3 × 4 + 0 = 12

At (0, 4), Z = 3x + 4y = 0 + 4 × 4 = 16

At (0, 0), Z = 0.

∴ Maximum value of Z = 16.

Question 2.

Minimise Z = – 3x + 4y

subject to x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0.

Solution:

Objective function Z = – 3x + 4y

Constraints are x + 2y ≤ 8, 3x + 2y ≤ 12, x ≥ 0, y ≥ 0

(i) Consider the line

x + 2y = 8

It passes through A(8, 0) and B(0, 4).

Putting x = 0, y = 0 in

x + 2y ≤ 8, 0 ≤ 8 which is true.

⇒ region x + 2y ≤ 8 lies on and below AB.

(ii) The line 3x + 2y = 12 passes through P(4, 0), Q(0, 6).

Putting x = 0, y = 0 in 3x + 2y ≤ 12.

⇒ 0 ≤ 12, which is true.

∴ Region 3x + 2y ≤ 12 lies on and below PQ.

(iii) x ≥ 0, the region lies on and to the right of y-axis.

(iv) y ≥ 0 lies on and above x-axis.

(v) Solving the equations x + 2y = 8 and 3x + 2y = 12,

we get x = 2, y = 3

⇒ R is (2,3), where AB and PQ intersect.

The shaded region OPRB is the feasible region.

At P(4, 0), Z = – 3x + 4y = – 12 + 0 = – 12

At R(2, 3), Z = – 6 + 12 = 6

At B(0, 4), Z = 0 + 16 = 16

At Q(0, 0), Z = 0

Thus, minimum value of Z is – 12 at P(4, 0) [Neglecting the value at (0, 0)].

Question 3.

Maximise Z = 5x + 3y.

subject to 3x + 5y ≤ 15, 5x + 2y ≤ 10, x ≥ 0, y ≥ 0.

Solution:

The objective function is Z = 5x + 3y.

Constraints are Y

3x + 5y ≤ 15,

5x + 2y ≤ 10, x ≥ 0, y ≥ 0

(i) Consider the line 3x + 5y = 15, which passes through A(5, 0) and B(0, 3).

Putting x = 0, y = 0, in 3x + 5y ≤ 15

⇒ 0 ≤ 15 which is true.

∴ Region 3x + 5y ≤ 15 lies on and below AB.

(ii) The line 5x + 2y = 10 passes through P(2, 0) and Q(0, 5).

Put x = 0, y = 0 in 5x + 2y ≤ 10

∴ 0 ≤ 10, which is true.

Region 5x + 2y ≤ 10 lies on and below PQ.

(iii) x ≥ 0 region lies on and to the right of y-axis.

(iv) y ≥ 0 lies on and above x-axis.

(v) The feasible region is the shaded area OPRB.

Solving the equations, 3x + 5y – 15 and 5x + 2y = 10,

we get x = \(\frac{20}{19}\), y = \(\frac{45}{19}\).

⇒ AB and PQ intersect at R (\(\frac{20}{19}\), \(\frac{45}{19}\)).

At P(2, 0), Z = 5x + 3y = 10 + 0 = 10

At R(\(\frac{20}{19}\), \(\frac{45}{19}\)),

Z = \(\frac{100}{19}\) + \(\frac{135}{19}\) = \(\frac{235}{19}\)

At B(0, 3), Z = 9

At O(0, 0), Z = 0

∴ Maximum value of Z = \(\frac{235}{19}\) at (\(\frac{20}{19}\), \(\frac{45}{19}\)).

Question 4.

Minimise P = 3x + 5y

subject to x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

Solution:

The objective function is Z = 3x + 5y.

Constraints are x + 3y ≥ 3, x + y ≥ 2, x, y ≥ 0.

(i) consider the line

x + 3y = 3.

This passes through A(3, 0) and B(0, 1).

Putting x = 0, y = 0 in x + 3y > 3

⇒ 0 ≥ 3, which is not true.

Region x + 3y ≥ 3 lies on and above AB.

(ii) The line x + y = 2 passes through C(2, 0) and D(0, 2)

Putting x = 0, y = 0 in x + y ≥ 2

⇒ 0 ≥ 2, which is not true.

⇒ Region x + y ≥ 2 lies on and above the line CD.

(iii) x ≥ 0, this region is on and to the right of y-axis.

(iv) y ≥ 0 is on and above the x-axis.

(v) The lines x + 3y = 3 and x + y = 2 intersect at R(\(\frac{3}{2}\), \(\frac{1}{2}\)).

(vi) The feasible region is XARDY.

At A(3, 0), Z = 3x + 5y = 9 + 0 = 9.

At R(\(\frac{3}{2}\), \(\frac{1}{2}\)), Z = \(\frac{9}{2}\) + \(\frac{5}{2}\) = \(\frac{14}{2}\) = 7.

At D(0, 2), Z = 0 + 5 × 2 = 10.

∴ Minimum value of Z at R(\(\frac{3}{2}\), \(\frac{1}{2}\)) is 7.

The feasible region XARDY is unbounded. Consider the inequality 3x + 5y < 7

The line 3x + 5y = 7 passes through (\(\frac{7}{3}\), 0) and (0, \(\frac{7}{5}\)).

3x + 5y < 7 has no common point with feasible region

∴ Minimum value of Z = 7 at R(\(\frac{3}{2}\), \(\frac{1}{2}\)).

Question 5.

Maximise Z = 3x + 2y

subject to x + 2y ≥ 10, 3x + y ≤ 15, x, y ≥ 0.

Solution:

The objective function is Z = 3x + 2 y.

Constraints are x + 2y ≤ 10,

3x + y ≤ 15,

x, y ≥ 0

(i) Consider the line

x + 2 y = 10.

This line passes through A(10, 0) and B(0, 5).

Putting x = 0, y = 0 in x + 2y ≤ 10

0 ≤ 10, which is true.

⇒ x + 2y ≤ 10 lies on and below AB.

(ii) The line 3x + y = 15 passes through P(5, 0) and Q(0, 15).

Putting x = 0, y = 0 is 3x + y ≤ 15, we get 0 ≤ 15, which is true.

⇒ 3x + y ≤ 15 lies on and below PQ.

(iii) x ≥ 0 lies on and to the right of y-axis.

(iv) y ≥ 0 lies on and above x-axis.

(v) The feasible region is OPRB.

(vi) Equations are

x + 2 y = 10 …………….. (1)

3x + y = 15 ………………… (2)

Multiplying (2) by 2 to obtains

6x + 2y – 30 ……………. (3)

Subtracting (1) from (3),

5x – 20

⇒ x = 4

From (2), y = 15 – 3x = 15 -12 = 3.

Lines (1) and (2) meet at R(4, 3).

Z = 3x + 2y

Now, At P(5, 0) Z = 15 + 0 = 15

At R(4, 3) Z = 12 + 6 = 18

At B(0, 5) Z = 0 + 10 = 10

∴ Maximum value of Z = 18 at R(4, 3).

Question 6.

Minimise Z = x + 2y

subject to 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

Show that minimum of Z occurs at more than two points.

Solution:

Objective function Z = x + 2y.

Constraints are 2x + y ≥ 3, x + 2y ≥ 6, x, y ≥ 0.

(i) Consider the line

2x + y = 3.

This passes through A(\(\frac{3}{2}\), 0) and B(0, 3).

Putting x = 0, y = 0 in 2x + y ≥ 3. we get 0 ≥ 3, which is not true.

∴ The region 2x + y ≥ 3 is on and above AB.

(ii) The line x + 2y = 6 passes through P(6, 0) and B(0, 3).

Putting x = 0, y = 0 in x + 2y ≥ 6

⇒ 0 ≥ 6 which is not true.

∴ x + 2y ≥ 6 region lies on PB and above it.

(iii) x ≥ 0 lies on and to the right of y-axis.

(iv) y ≥ 0 lies on and above x-axis.

Shaded area XPBY is the feasible region.

At P(6, 0), Z = x + 2y = 6 + 0 = 6.

At B(0, 3), Z = 0 = 6 = 6

⇒ At all points on PB, Z = 6

Moreover x + 2y < 6 lies below PB which has no common point in feasible region.

Thus, the minimum value of Z is 6.

Question 7.

Minimise and Maximise Z = 5x + 10y

subject to x + 2y ≤ 120, x + y ≥ 60, x – 2y ≥ 0, x, y ≥ 0.

Solution:

The objective functions is

Z = 5x + 10y.

Constraints are

x + 2y ≤ 120,

x + y ≥ 60,

x – 2y ≥ 0,

x, y ≥ 0

(i) The line x + 2y = 120 passes through A(120, 0) and 5(0, 60)

Putting x = 0, y = 0 is x + 2y ≤ 120,

we get 0 ≤ 120, which is true.

⇒ x + 2y ≤ 120 lies on AB and below AB.

(ii) The line x + y = 60 passes through P(60, 0) and B(0, 60).

Putting x = 0 and y = 0 in x + y ≥ 60, which is not true.

⇒ x + y ≥ 60 lies on PB or above it.

(iii) The line x – 2y = 0 passes through O(0, 0) and Q(120, 60).

Putting x = 60, y = 0 in x – 2y ≥ 0, 60 ≥ 0, which is true.

⇒ x – 2y = 0 region is on QQ and below it.

(iv) x ≥ 0 lies on y-axis on its right.

(v) y ≥ 0 lies on x-axis and above it.

⇒ Feasible region is PARS which has been shaded.

Question 8.

Minimise and Maximise Z = x + 2y

subject to x + 2y ≥ 100, 2x – y ≤ 0, 2x + y ≤ 200, x, y ≥ 0.

Solution:

Objective functions is Z = x + 2y.

Constraints are

x + 2y ≥ 100,

2x – y ≤ 0,

2x + y ≤ 200,

x, y ≥ 0

(i) The line x + 2y – 100, passes through A(100, 0) and B(0, 50).

Putting x = 0, y = 0 in x + y ≥ 100 or 0 ≥ 100, which is not true.

∴ Region x + 2y ≥ 100 lies on and above AB.

(ii) The line 2x – y = 0 passes through O(0, 0) and C(50, 100).

The point (0, 50) lies in the region 2x – y ≤ 0

∴ Region 2x – y ≤ 0 lies on and above OC.

(iii) The line 2x + y = 200 passes through A(100, 0) and D(0, 200).

Put x = 0, y = 0 in 2x + y ≤ 200, i.e., 0 ≤ 200, which is true.

⇒ 2x + y ≤ 200 lies on AD and below it.

(iv) x ≥ 0 lies on y-axis and on the right of it.

(v) y ≥ 0 lies on and above x-axis.

Feasible region is the shaded area BECD.

The point E is the intersection of

OC : 2x – y = 0 and AB : x + 2y = 100

i.e., The point E is (20, 40).

The point C(50, 100) lies on 2x + y = 200 and 2x – y = 0.

Now,

At B(0,50), Z = x + 2y = 100

At E(20, 40), Z = 20 + 80 = 100

At C(50, 100), Z = 50 + 200 = 250

At D(0, 200), Z = 0 + 2 × 200 = 400

Minimum value of Z = 100 at all points on the line segment joining the points B(0, 50) and E(20, 40).

Maximum value of Z = 400 at D(0, 200).

Question 9.

Maximise Z = – x + 2y, subject to constraints:

x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0.

Solution:

The objective function is Z = – x + 2y. The constraints are:

x ≥ 3, x + y ≥ 5, x + 2y ≥ 6, y ≥ 0

(i) The line x + y = 5 passes through A(5, 0), B(0, 5).

Putting x = 0, y = 0 in x + y ≥ 5, we get 0 ≥ 5, which is not true.

⇒ x + y ≥ 5 lies on and above AB.

(ii) The line x + 2y = 6 passes through C(6, 0), D(0, 3).

Putting x = 0, y = 0 in x + 2y ≥ 6,

we get 0 ≥ 6, which is not true.

⇒ x + 2y ≥ 6 lies on and above CD.

(iii) x ≥ 3 lies on PQ or on the right of it.

(iv) y ≥ 0 lies on and above x-axis.

(v) The feasible region is PQRCX.

(a) Solving x = 3 and x + y = 5, we get x = 3, y = 2.

∴ These lines meet at Q(3, 2).

(b) x + y = 5 and x + 2y = 6 meet at R,

i.e., the point R is (4, 1).

At Q(3, 2), Z = – x + 2y = – 3 + 4 = 1

At R(4, 1), Z = – 4 + 2 = – 2

At C(6, 0), Z = – 6 + 0 = – 6

The maximum value of Z is 1 but the feasible region is unbounded.

Consider the inequality – x + 2y > 1

The line – x + 2y = 1 passes through (- 1, 0) and (0, \(\frac{1}{2}\)).

Putting x = 0, y = 0 in – x + 2y > 1, we get 0 > 1, which is not true.

⇒ – x + 2y > 1 lies above the line – x + 2y = 1.

∴ Feasible region of – x + 2y > 1 have many points in common.

Therefore, there is no maximum value.

Question 10.

Maximise Z = x + y subject to

x – y ≤ – 1, – x + y ≤ 0, x, y ≥ 0.

Solution:

Objective function is Z = x + y.

Constraints are x – y ≤ – 1,

– x + y ≤ 0, x, y ≥ 0.

(i) The line x – y = – 1 passes through (- 1, 0) and (0, 1).

Putting x = 0, y = 0 in x – y ≤ – 1,

we get 0 ≤ – 1 which is not true.

⇒ x – y ≤ – 1 lies on and above

AB : x – y = – 1.

(ii) the line – x + y = 0 passes through O(0, 0) and C(1, 1).

Putting x = 0, y = 1 in – x + y ≤ 0,

we get 1 ≤ 0, which is not true.

⇒ – x + y ≤ 0 lies on and below OC.

(iii) x ≥ 0 lies on and on the right of y-axis.

(iv) y ≥ 0 lies on and above x-axis.

There is no common region, i.e., their is no feasible region.

⇒ There is no maximum value of Z.