Ex 6.3, 14 - Chapter 6 Class 11 Linear Inequalities (Term 2)
Last updated at Feb. 17, 2020 by Teachoo
Ex 6.3
Last updated at Feb. 17, 2020 by Teachoo
Ex 6.3, 14 Solve the following system of inequalities graphically: 3x + 2y ≤ 150, x + 4y ≤ 80, x ≤ 15, y ≥ 0, x ≥ 0 Now we solve 3x + 2y ≤ 150 Lets first draw graph of 3x + 2y = 150 Putting x = 0 in (1) 3(0) + 2y = 150 0 + 2y = 150 2y = 150 y = 150/2 y = 75 Putting y = 0 in (1) 3x + 2(0) = 150 3x + 0 = 150 3x = 150 x = 150/3 x = 50 Points to be plotted are (0,75) , (50,0) Drawing graph Checking for (0,0) Putting x = 0, y = 0 3x + 2y ≤ 150 4(0)+ 3(0)≤ 150 0 ≤ 150 which is true Hence origin lies in plane 3x + 2y ≤ 150 So, we shade left side of line Now we solve x + 4y ≤ 80 Lets first draw graph of x + 4y = 80 Putting x = 0 in (2) 0 + 4y = 80 4y = 80 y = 0a y = 20 Putting y = 0 in (2) x + 4(0) = 80 x + 0 = 80 x = 80 Points to be plotted are (0,20) , (80,0) Drawing graph Checking for (0,0) Putting x = 0, y = 0 x + y ≤ 80 0 + y(0) ≤ 80 0 ≤ 80 which is true Hence (0, 0) lies in plane x + 4y ≤ 80 So, we shade left side of line Also, x ≤ 15 So, for all values of y, x = 15 Given x ≤ 15, so we shade left side of line Also, given x, y ≥ 0 So, the shaded region will lie in 1st quadrant. Hence the shaded region represents the given inequality.