Chapter 12: Linear Programming

Let x units of type A and y units of type B electric circuits be produced by the manufacturer.

From the given information the below table is constructed:

Now, the total profit function in rupees Z = 50x + 60y is to be maximized with subject to the constraints

20x + 10y ≤ 200 … (i); 10x + 20y ≤ 120 … (ii)

10x + 30y ≤ 150 … (iii); x ≥ 0, y ≥ 0 … (iv)

Therefore, the required LPP is

Maximize Z = 50x + 60y subject to the constraints

20x + 10y ≤ 200 2x + y ≤ 20;

10x + 20y ≤ 120 x + 2y ≤ 12 and

10x + 30y ≤ 150 x + 3y ≤ 15, x ≥ 0, y ≥ 0.

Let’s consider x and y to be the number of large and small vans respectively.

From the given information the below constrains table is constructed:

Now, the objective function for minimum cost is

Z = 400x + 200y

Subject to the constrains;

200x + 80y ≥ 1200 ⇒ 5x + 2y ≥ 30 ….. (i)

400x + 200y ≤ 3000 ⇒ 2x + y ≤ 15 ….. (ii)

x ≤ y …. (iii)

and x ≥ 0, y ≥ 0 (non-negative constraints)

Therefore, the required LPP is to minimize Z = 400x + 200y

Subject to the constraints 5x + 2y ≥ 30, 2x + y ≤ 15, x ≤ y and x ≥ 0, y ≥ 0.

Let’s consider that the company manufactures x boxes of type A screws and y boxes of type B screws.

From the given information the below table is constructed:

From the data in the above table, the objective function for maximum profit Z = 100x + 170y

Subject to the constraints

2x + 8y ≤ 3600 ⇒ x + 4y ≤ 1800 … (i)

3x + 2y ≤ 3600 … (ii)

x ≥ 0, y ≥ 0 (non-negative constraints)

Therefore, the required LPP is

Maximize: Z = 100x + 170y

Subject to constraints,

x + 4y ≤ 1800, 3x + 2y ≤ 3600, x ≥ 0, y ≥ 0.

Let’s assume x and y to be the number of sweaters of type A and type B respectively.

From the question, the following constraints are:

360x + 120y ≤ 72000 ⇒ 3x + y ≤ 600 … (i)

x + y ≤ 300 … (ii)

x + 100 ≥ y ⇒ y ≤ x + 100 … (iii)

Profit: Z = 200x + 120y

Therefore, the required LPP to maximize the profit is

Maximize Z = 200x + 120y subject to constrains

3x + y ≤ 600, x + y ≤ 300, y ≤ x + 100, x ≥ 0, y ≥ 0.

Let’s assume the man covers x km on his motorcycle at the speed of 50km/hr and covers y km at the speed of 50 km/hr and covers y km at the speed of 80 km/hr.

So, cost of petrol = 2x + 3y

The man has to spend Rs 120 atmost on petrol

⇒ 2x + 3y ≤ 120 …. (i)

Now, the man has only 1 hr time

So, x/50 + y/80 ≤ 1 ⇒ 8x + 5y ≤ 400 … (ii)

And, x ≥ 0, y ≥ 0

To have maximum distance Z = x + y.

Therefore, the required LPP to travel maximum distance is maximize Z = x + y, subject to the constraints

2x + 3y ≤ 120, 8x + 5y ≤ 400, x ≥ 0, y ≥ 0.

Long Answer (L.A.)

As per the solution of exercise 11, we have

Maximize Z = 50x + 60y subject to the constraints

20x + 10y ≤ 200 2x + y ≤ 20 … (i)

10x + 20y ≤ 120 x + 2y ≤ 12 … (ii)

10x + 30y ≤ 150 x + 3y ≤ 15 … (iv)

x ≥ 0, y ≥ 0 … (iv)

Now, let’s construct a constrain table for the above

Table for (i)

Table for (ii)

Table for (iii)

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

x = 28/3, y = 4/3

So, the corner point is B(28/3, 4/3).

Solving equations (ii) and (iii) we get,

x = 6, y = 3 and the corner point is C(6, 3)

Lastly, solving equations (i) and (iii) we get,

x = 9, y = 2 (not included in the feasible region)

Here, OABCD is the feasible region.

Hence, the corner points are O(0, 0), A(10, 0), B(28/3, 4/3), C(6, 3) and D(0, 5).

Let us evaluate the value of Z

So here, the maximum profit is Rs 546.6 which is not possible for number of items in fraction.

Therefore, the maximum profit for the manufacture is Rs 480 at (6, 3) i.e. Type A = 6 and Type B = 3.

As per the solution of exercise 12, we have

The objective function for minimum cost is Z = 400x + 200y

Subject to the constrains;

5x + 2y ≥ 30 ….. (i)

2x + y ≤ 15 ….. (ii)

x ≤ y …. (iii)

and x ≥ 0, y ≥ 0 (non-negative constraints)

Now, let’s construct a constrain table for the above

Table for (i)

Table for (ii)

Table for (iii)

Next, solving equations (i) and (iii), we get

x = 30/7 and y = 30/7, so the corner point is A(30/7, 30/7)

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

x = 5 and y = 5, so the corner point is B(5, 5)

Here, ABC is the shaded feasible region whose corner points are A(30/7, 30/7), B(5, 5) and C(0, 15)

On evaluating the value of Z, we have

From the table it’s seen that the minimum value is 2571.4

Therefore, the required minimum cost is Rs 2571.4 at (30/7, 30/7)

As per the solution of exercise 13, we have

The objective function for maximum profit Z = 100x + 170y

Subject to constraints,

x + 4y ≤ 1800 …. (i)

3x + 2y ≤ 3600 … (ii)

x ≥ 0, y ≥ 0

Now, let’s construct a constrain table for the above

Table for (i)

Table for (ii)

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

x = 1080 and y = 180

It’s seen that OABC is the feasible region whose corner points are O(0, 0), A(1200, 0), B(1080, 180) and C(0, 450).

On evaluating the value of Z, we have

Form the table it’s seen that the maximum value of Z is 138600.

Therefore, the maximum profit of the function Z is 138600 at (1080, 180).

As per the solution of exercise 14, we have

Maximize Z = 200x + 120y subject to constrains

3x + y ≤ 600 …. (i)

x + y ≤ 300 …. (ii)

x – y ≤ -100 …. (iii)

x ≥ 0, y ≥ 0

Now, let’s construct a constrain table for the above

Table for (i)

Table for (ii)

Table for (iii)

Next, solving equation (i) and (iii) we get

x = 100 and y = 200

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

x = 150 and y = 150

It’s seen that the shaded region is the feasible region whose corner points are O(0, 0), A(200, 0), B(150, 150), D(0, 100).

Evaluating the value of Z, we have

From the above table it’s seen that the maximum value is 48000.

Therefore, the maximum value of Z is 48000 at (150, 150) which means 150 sweaters of each type.

As per the solution of exercise 15, we have

Maximize Z = x + y, subject to the constraints

2x + 3y ≤ 120 … (i)

8x + 5y ≤ 400 … (ii)

x ≥ 0, y ≥ 0

Now, let’s construct a constrain table for the above

Table for (i)

Table for (ii)

Next, solving equation (i) and (iii) we get

x = 300/7 and y = 80/7

It’s seen that the feasible region whose corner points are O(0, 0), A(50, 0), B(300/7, 80/7) and C(0, 40).

Let’s evaluate the value of Z

From above table the maximum value of Z is 54.3

Therefore, the maximum distance that the man can travel is 54.3 km at (300/7, 80/7).