Page 1 : EXERCISE 12.1 PAGE NO: 513, 1. Maximise Z = 3x + 4y, , Subject to the constraints: **” <4." = 0720., , Solution:, , The feasible region determined by the constraints, x + y = 4, x 20, y 20, is given below, , ees =, , , , ‘ xey=4, 0 (0, 0), A (4, 0), and B (0, 4) are the corner points of the feasible region. The values of, Z at these points are given below, , , , , , , , , , , , , , , , , , Corner point Z=3x+4y, , 00,0) 0, , A (4, 0) 12, , B (0, 4) 16 Maximum, Hence, the maximum value of Z is 16 at the point B (0, 4), 2. Minimise Z = —3x + 4y, subject to*+2yS8 3x42ysl2, x20, y20_, Solution:, , The feasible region determined by the system of constraints,, x42y 28 3r4¢2y<12, x20, y20 is given below

Page 2 : ax42y~12, O (0, 0), A (4, 0), B (2, 3) and C (0, 4) are the corner points of the feasible region, The values of Z at these corner points are given below, , , , , , , , , , , , , , , , , , , , Corner point Z=-3x+4y, , | 0 (0, 0) a eS 2 o, A(4,0) “12 1 Minimum, B (2, 3) __6 Lo, C(0, 4) 16, , Hence, the minimum value of Z is — 12 at the point (4, 0), , 3. Maximise Z = Sx + 3y, , subject to3* +59 515. Sx+2y 510, x20, y20,, , Solution:, , The feasible region determined by the system of constraints, 3x + Sy < 15, Sx + 2y <, , , , n+ Sy=lS, , : 5x4 2y=10, , O (0, 0), A (2, 0), B (0, 3) and C (20 / 19, 45 / 19) are the corner points of the feasible

Page 3 : region. The values of Z at these comer points are given below, , , , , , , , , , , , , , , , , , , , , , Comer point Z=5x + 3y, , 0 (0,0) 0, , A(, 0) 10, , B (0, 3) 9, L C (20/19, 45/19) 235/19 Maximum, Hence, the maximum value of Z is 235 / 19 at the point (20 / 19, 45 / 19), 4. Minimise Z = 3x + Sy, such that **3¥ 23, e+ 22, xy20,, Solution:, , The feasible region determined by the system of constraints, x + 3y 2 3, x + y > 2, and x, y, = Ois given below, , , , It can be seen that the feasible region is unbounded., The corner points of the feasible region are A (3, 0), B (3 / 2, 1/2) and C (0, 2), The values of Z at these comer points are given below, , Comer Z=3x+5, AGO 9, , BG/2, 1/2 7 Smallest, CO,2 10, , , , 7 may or may not be the minimum value of Z because the feasible region is unbounded, For this purpose, we draw the graph of the inequality, 3x + Sy < 7 and check the resulting, half plane have common points with the feasible region or not, , Hence, it can be seen that the feasible region has no common point with 3x + Sy <7

Page 4 : Thus, the minimum value of Z is 7 at point B (3 / 2, 1/2), , 5. Maximise Z = 3x + 2y, , subject to* +2" 510.504) SI5,x,y20,, , Solution:, , The feasible region determined by the constraints, x + 2y < 10, 34 + y < 15, x20, and y>, 0, is given below, , vy?, , , , . ede, , , , The values of Z at these corner are given below, , , , , , Z=3x+2y, , , , 15, , , , , , , , , , , , , , , , , , Hence, the maximum value of Z is 18 at the point (4, 3), , 6. Minimise Z = x + 2y, , subject to2* + = 3. x+2y26,x,y20°, , Solution:, , The feasible region determined by the constraints, 2x + y > 3,.x + 2y > 6, x>0, and y> 0,, is given below

Page 5 : tyes x+ay=6, , A (6, 0) and B (0, 3) are the corner points of the feasible region, The values of Z at the corner points are given below, , , , , , , , , , , , Corner point Z=x+2y, A(6, 0) 6, | B (0, 3) _ 6, Here, the values of Z at points A and B are same. If we take any other point such as (2, 2), on line x + 2y = 6, then Z=6, , Hence, the minimum value of Z occurs for more than 2 points,, Therefore, the value of Z is minimum at every point on the line x + 2y = 6, , 7. Minimise and Maximise Z = Sx + 10y, , subject to* +2) $120,x+ y 2 60,x~-2y20,x,y20,, , Solution:, , The feasible region determined by the constraints, x + 2y < 120, x + y= 60, .« -2y >, 0, x= 0, and y > 0, is given below