Question

8 Minimize z= x + 3y 9 + 22 54 + 4yΣ Subject to 2y + 2 > ΛΙ ΛΙ ΛΙΛΙ ΛΙ 14 O Σ Ο Minimum is

Answer 1

=), which passes through (80) So this also not true. Given, 2 = 7+37 Minimize subject to 4+256) 8 57 +44), 32 24 + 2H), 14 Ho 40. now, we draw the line Y+22=8 20+ y = 8 Ales) 들 1봉에 AB (1,6) (54) This passes through (4,0), (0,8) If we put 60) in Y+24,8, we get oto78 which is not true so the solution must be opposite to origin, and, 54. UR=32 nt 32 gives 4R+ 5y = 32 y 8 And 0+0), 32, solution The then region is opposite to origin. (5)

Now we draw 2x+2y=14, -), which passes through n t =) 7 7 (7,0), (0,7) 2-0 + 2:0 208,14. so is also opposite to solution this region origen Hence the solution region is ABCD unbounde region, where, A= (0,8), B= (196), C= (54) and D(8,0). Mowe, z]A (+34 39) Icones - Ot24 224. ZB [1434 (6) B - 1 + 18-19 5+ 3xy Z) ( (54) = zocro 5+12-17 Hence 8+300 = 8 z is minimum at n=8,9 = 0 and Zmin = 8