Practicing Success
$z = 30x+20y, x+y ≤ 8, x + 2y ≥ 4, 6x + 4y ≥ 12, x≥0, y ≥ 0$ has |
Unique solution Infinitely many solution Minimum at (4, 0) Minimum 60 at point (0, 3) |
Minimum 60 at point (0, 3) |
Since, $x+y=8$ ....(i) This line meets axes at (8,0) and (0,8) respectively. $x+2y=4$ .....(ii) $⇒\frac{x}{4}+\frac{y}{2}=1$ This line meet axes at (4,0) and (0,2). And $6x+4y=12$ ......(iii) $⇒\frac{x}{2}+\frac{y}{3}=1$ This line meets axes at (2,0) and (0,3) The point of intersection of equations (ii) and (iii) is $F(1,\frac{3}{2})$ Now, at A(4,0), $z=30×4=120$ B(8,0), $z=30×8=240$ C(0,8), $z=20×8=160$ D(0,3), $z=20×3=60$ and $F(1,\frac{3}{2}),z=30×1+20×\frac{3}{2}=60$ It is clear that z is minimum 60 at points D(0,3) and $F(1,\frac{3}{2})$ |