RD Sharma Solutions for Class 12 Maths Chapter 30 Linear Programming Exercise 30.4, is given here to help students excel in their board exams. The solutions are designed by our subject experts, to help understand the concepts and methods to solve problems in a shorter period, and also to boost the confidence levels among students. Download the RD Sharma Class 12 Solutions PDF for free from the links given below.
Download PDF of RD Sharma Solutions for Class 12 Maths Chapter 30 Exercise 4
Access RD Sharma Solutions for Class 12 Maths Chapter 30 Exercise 4
EXERCISE 30.4
Q1.
Solution:
Let us consider a young man drive x km at a speed of 25km/hr and y km at a speed of 40km/hr.
Let Z be the total distance travelled by young man.
So, Z = x + y
First constraint:
When speed is 25km/hr, the young man spends = Rs 2 per km
When speed is 40km/hr, the young man spends = Rs 5 per km
So,
Expenses on x km and y km = Rs 2x and Rs 5y
But young man has only Rs 100
So, 2x + 5y ≤ 100
Second constraint:
Time taken to travel x km = Distance/speed
= x/25 hr
Time taken to travel y km = Distance/speed
= y/40 hr
It is given that, 1hr to travel,
So, x/25 + y/40 ≤ 1
40x + 25y ≤ 1000
8x + 5y ≤ 200
Hence, the required mathematical formulation of linear programming is:
Minimize Z = x + y
Subject to constraints
2x + 5y ≤ 100
8x + 5y ≤ 200
Where, x, y ≥ 0
Now, region 2x + 5y ≤ 100:
The line 2x + 5y = 100 meets axes at A1(50, 0), B1(0, 20) region containing origin represents 2x + 5y ≤ 100 as (0, 0) satisfies 2x + 5y ≤ 100.
Region 8x + 5y ≤ 200:
The line 8x + 5y = 200 meets axes at A2(25, 0), B2(0, 40) region containing origin represents 8x + 5y ≤ 200 as (0, 0) satisfies 8x + 5y ≤ 200.
Region x, y ≥ 0: it represents first quadrant in xy-plane.
Shaded region O A2 P B1 represents feasible region.
Point P(50/3, 40/3) is obtained by solving 8x + 5y = 200, 2x + y = 100
The value of Z = x + y at
O(0, 0) = 0 + 0 = 0
A2(25, 0) = 25 + 0 = 25
P(50/3, 40/3) = 50/3 + 40/3 = 30
B1(0, 20) = 0 + 20 = 20
Hence, maximum value of Z = 30 at x = 50/3, y = 40/3
Distance travelled at speed of 25km/hr = 50/3 km
And at speed of 40 km/hr = 40/3 km
So maximum distance = 30 km
Q2.
Solution:
Let us consider the required quantity of items be A and B.
Given:
Profits on one item A and B = Rs 6 and Rs 4
So, profits on x items of type A and y items on type B = 6x and 4y
Let the total profit be Z,
Z = 6x + 4y
First constraint:
Machine I works 1 hour and 2hours on item A and B
So, x no. of item A and y no. of item B = x hour and 2y hours on machine I
Machine I works at most = 12 hours
x + 2y ≥ 12
Second constraint:
Machine II works 2 hours and 1hour on item A and B
So, x no. of item A and y no. of item B = 2x hours and y hours on machine II
Machine II works maximum of = 12 hours
2x + y ≥ 12
Third constraint:
Machine III works 1 hour and 5/4hour on item A and B
So, x no. of item A and y no. of item B = x hours and 5/4y hours on machine III
Machine III works at least = 5 hours
x + 5/4y ≥ 5
4x + 5y ≥ 20
Hence, the required mathematical formulation of linear programming is:
Minimize Z = 6x + 4y
Subject to constraints
x + 2y ≥ 12
2x + y ≥ 12
4x + 5y ≥ 20
Where, x, y ≥ 0
Now, region x + 2y ≥ 12:
The line x + 2y = 12 meets axes at A1(12, 0), B1(0, 6) region containing origin represents x + 2y ≥ 12 as (0, 0) satisfies x + 2y ≥ 12.
Region 2x + y ≥ 12:
The line 2x + y = 12 meets axes at A1(6, 0), B1(0, 12) region containing origin represents 2x + y ≥ 12 as (0, 0) satisfies 2x + y ≥ 12.
Region 4x + 5y ≥ 20:
The line 4x + 5y = 20 meets axes at A3(5, 0), B3(0, 4) region not containing origin represents 4x + 5y ≥ 20 as (0, 0) does not satisfy 4x + 5y ≥ 20.
Region x, y ≥ 0: it represents first quadrant in xy-plane.
Shaded region A2 A3 P B3 B1 represents feasible region.
The value of Z = 6x + 4y at
A2(6, 0) = 6(6) + 4(0) = 36
A3(5, 0) = 6(5) + 4(0) = 30
B3(0, 4) = 6(0) + 4(4) = 16
B2(0, 6) = 6(0) + 4(6) = 24
P(4,4) = 6(4) + 4(4) = 40
Hence, maximum value of Z = 40 at x = 4, y = 4
Required number of product A =4, product B = 4
So maximum profit = Rs 40
Q3.
Solution:
Let us consider tailor A and B work for x and y days.
Given:
Tailor A and B earn = Rs 15 and Rs 20
So, tailor A and B earn is x and y days = Rs 15x and 20y
Let the total maximum profit be Z which gives minimum labour cost,
Z = 15x + 20y
First constraint:
Tailor A and B stitches = 6 and 10 shirts in a day
So, tailor A and B can stitch = 6x and 10y shirts in a day
But, tries to produce atleast = 60 shirts
So, 6x + 10y ≥ 60
3x + 5y ≥ 30
Second constraint:
Tailor A and B stitches = 4 pants each in a day
So, tailor A and B can stitch = 4x and 4y pants in a day
But, tries to produce atleast = 32 pants
So, 4x + 4y ≥ 32
x + y ≥ 8
Hence, the required mathematical formulation of linear programming is:
Minimize Z = 15x + 20y
Subject to constraints
3x + 5y ≥ 30
x + y ≥ 8
Where, x, y ≥ 0
Now, region 3x + 5y ≥ 30:
The line 3x + 5y = 30 meets axes at A1(10, 0), B1(0, 6) region not containing origin represents 3x + 5y ≥ 30 as (0, 0) does not satisfy 3x + 5y ≥ 30.
Region x + y ≥ 8:
The line x + y = 8 meets axes at A2(8, 0), B2(0, 8) region not containing origin represents x + y ≥ 8 as (0, 0) does not satisfy x + y ≥ 8.
Region x, y ≥ 0: it represents first quadrant in xy-plane.
Unbounded shaded region A1 P B2 represents feasible region with corner points A1(10, 0), P(5, 3), B2(0, 8)
The value of Z = 15x + 20y at
A1(10, 0) = 15(10) + 20(0) = 150
P(5, 3) = 15(5) + 20(3) = 135
B2(0, 8) = 15(0) + 20(8) = 160
Hence, smallest value of Z = 135
Open half plane 15x + 20y < 135 has no point in common with feasible region
So, smallest value is the minimum value.
Z = 135, at x = 5, y = 3
Therefore, tailor A should work for 5 days and B should work for 3 days.
Q4.
Solution:
Let us consider the factory manufactures x screws of type A and y screws of type B per day.
So, x ≥ 0 and y ≥ 0
| Screw A | Screw B | Availability | |
| Automatic Machine (min) | 4 | 6 | 4 x 60 = 120 |
| Hand operated Machine (min) | 6 | 3 | 4 x 60 = 120 |
The profit on package of screws A and screws B is Rs 7 and Rs 10.
So the constraints are:
4x + 6y ≤ 240
6x + 3y ≤ 240
Total profit, Z = 7x + 10y
Hence, the required mathematical formulation of linear programming is:
Minimize Z = 7x + 10y
Subject to constraints
4x + 6y ≤ 240
6x + 3y ≤ 240
Where, x, y ≥ 0
The feasible region obtained by the constraints is:
The corner points are A(40, 0), B(30, 20) and C(0, 40)
So the value of Z at these corner points is:
| Corner point | Z = 7x + 10y | |
| A(40, 0) | 280 | |
| B(30, 20) | 410 | maximum |
| C(0, 40) | 400 |
Hence, the maximum value of Z = 410 at B(30, 20).
So, the factory should produce 30 packages of screws A and 20 packages of screws B to get the maximum profit of Rs 410.
Q5.
Solution:
Let us consider the required number of belt A and belt B be x and y.
Given:
Profits on belt A and B = Rs 2 and Rs 1.50 per belt
So, profits on x belts of type A and y belts on type B = 2x and 1.5y
Let the total profit be Z,
Z = 2x + 1.5y
First constraint:
Each belt of type A requires twice as much as belt B.
Each belt of type B requires = 1 hour to make, so A requires = 2 hours
So for x and y belts of type A and B requires = 2x and y hours to make
Total time available is equal to production of 1000 belts B = 1000 hours
2x + y ≤ 1000
Second constraint:
It is given that, supply of leather for = 800 belts per day for both A and B
So, x + y ≤ 800
Third constraint:
It is given that, buckles available for A = 400 and
Buckles available for B = 700
So, x ≤ 400
y ≤ 700
Hence, the required mathematical formulation of linear programming is:
Maximize Z = 2x + 1.5y
Subject to constraints
2x + y ≤ 1000
x + y ≤ 800
x ≤ 400
y ≤ 700
Where, x, y ≥ 0
Now, region 2x + y ≤ 1000:
The line 2x + y = 1000 meets axes at A1(500, 0), B1(0, 1000) region containing origin represents 2x + y ≤ 1000 as (0, 0) satisfies 2x + y ≤ 1000.
Region x + y ≤ 800:
The line x + y = 800 meets axes at A2(800, 0), B2(0, 800) region containing origin represents x + y ≤ 800 as (0, 0) satisfies x + y ≤ 800.
Region x ≤ 400:
The line x ≤ 400 is parallel to y-axis and meets x-axis at A3(400, 0) region containing origin represents x ≤ 400 as (0, 0) satisfies x ≤ 400.
Region y ≤ 400:
The line y ≤ 700 is parallel to x-axis and meets y-axis at B3(0, 700) region containing origin represents y ≤ 700 as (0, 0) satisfies y ≤ 700.
Region x, y ≥ 0: it represents first quadrant in xy-plane.
Shaded region O A3 P Q R B3 represents feasible region.
Point P is obtained by intersection of x + y = 800, 2x + y = 1000,
R is not point of intersection of y = 700, x + y = 800.
The value of Z = 2x + 1.5y at
O(0, 0) = 2(0) + 1.5(0) = 0
A3(400, 0) = 2(400) + 1.5(0) = 800
P(400, 200) = 2(400) + 1.5(200) = 1100
Q(200, 600) = 2(200) + 1.5(600) = 1300
R(100, 700) = 2(100) + 1.5(700) = 1250
B3(0, 700) = 2(0) + 1.5(700) = 1050
Hence, maximum value of Z = 1300 at x = 200, y = 600
Required no. of belt A = 200,
Required no. of belt B = 600,
So maximum profit = Rs 1300
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