2 Evaluate the objective function Z ax by at each corner point Let M and m

2 evaluate the objective function z ax by at each

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(2) Evaluate the objective function Z = ax + by at each corner point. Let M and m , respectively denote the largest and the smallest values of Z. (3) (i) When the feasible region is bounded , M and m are, respectively, the maximum and minimum values of Z. (ii) In case, the feasible region is unbounded . (a) M is the maximum value of Z, if the open half plane determined by ax + by > M has no point in common with the feasible region. Otherwise, Z has no maximum value. (b) Similarly, m is the minimum of Z, if the open half plane determined by ax + by < m has no point in common with the feasible region. Otherwise, Z has no minimum value. 12.1.12 Multiple optimal points If two corner points of the feasible region are optimal solutions of the same type, i.e., both produce the same maximum or
minimum, then any point on the line segment joining these two points is also an optimal solution of the same type. 12.2 Solved Examples Short Answer (S.A.) Example 1 Determine the maximum value of Z = 4 x + 3 y if the feasible region for an LPP is shown in Fig. 12.1. LINEAR PROGRAMMING 243 Solution The feasible region is bounded. Therefore, maximum of Z must occur at the corner point of the feasible region (Fig. 12.1). (Maximum) Hence, the maximum value of Z is 112. Example 2Determine the minimum value of Z = 3x+ 2y(if any), if the feasibleregion for an LPP is shown in Fig.12.2. Corner Point Value of Z O, (0, 0) 4 (0) + 3 (0) = 0 A (25, 0) 4 (25) + 3 (0) = 100 B (16, 16) 4 (16) + 3 (16) = 112 C (0, 24) 4 (0) + 3 (24) = 72
(smallest) 244 MATHEMATICS Let us graph 3 x + 2 y < 13. We see that the open half plane determined by 3 x + 2 y < 13 and R do not have a common point. So, the smallest value 13 is the minimum value of Z. Example 3Solve the following LPP graphically:Maximise Z = 2x+ 3y, subject to x+ 4,x 0, y y 0 y 4. Corner Point Value of Z A, (12, 0) 3 (12) + 2 (0) = 36 B (4, 2) 3 (4) + 2 (2) = 16 C (1, 5) 3 (1) + 2 (5) = 13 D (0, 10) 3 (0) + 2 (10) = 20
LINEAR PROGRAMMING 245 Hence, the maximum value of Z is 12 at the point (0, 4)

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