Consider the linear programming problem(LPP):
Minimize $Z = x+y$
$x + 2y ≤ 4,$
$3x + y ≥ 3,$
$4x + 3y ≥ 6,$
$x, y ≥ 0$.
Which of the following is correct for the above linear programming problem (LPP):

(A) The LPP has a bounded feasible region.
(B) The LPP has a unique optimal solution.
(C) The optimal value of the LPP exists at the point (3/2, 0)
(D) The corner points of the feasible region are (3/2, 0), (3/5, 6/5), (2/5, 6/5) and (4, 0)

Choose the correct answer from the options given below:

(A), (B) and (C) only

The correct answer is Option (2) → (A), (B) and (C) only

Constraints:

$x + 2y \le 4$

$3x + y \ge 3$

$4x + 3y \ge 6$

$x, y \ge 0$

Corner points are obtained by solving the boundary line intersections:

(1) $x + 2y = 4$ and $3x + y = 3$

Solution: $(\frac{2}{5}, \frac{9}{5})$

(2) $3x + y = 3$ and $4x + 3y = 6$

Solution: $(\frac{3}{5}, \frac{6}{5})$

(3) Intersection with $y = 0$ gives feasible points:

$x \in [\frac{3}{2}, 4]$ → corner points: $(\frac{3}{2}, 0)$ and $(4, 0)$

Thus true corner points: $(\frac{3}{2}, 0), (4, 0), (\frac{3}{5}, \frac{6}{5}), (\frac{2}{5}, \frac{9}{5})$

Now evaluate the objective function $Z = x + y$ at each corner:

$Z(\frac{3}{2}, 0) = 1.5$

$Z(4, 0) = 4$

$Z(\frac{3}{5}, \frac{6}{5}) = \frac{9}{5} = 1.8$

$Z(\frac{2}{5}, \frac{9}{5}) = \frac{11}{5} = 2.2$

Minimum value is $1.5$ at the point $(\frac{3}{2}, 0)$ and is unique.

Analysis of statements:

(A) Feasible region is bounded. ✔

(B) Unique optimal solution exists. ✔

(C) Optimal value occurs at $(\frac{3}{2}, 0)$. ✔

(D) Corner points listed are incorrect. ✘

Correct options: (A), (B), (C)