Consider the linear programming problem (LPP): Maximize $Z = 6x + 3y$ subject to the conditions, $4x + y ≥80, x+5y ≥ 115, 3x + 2y ≤ 150, x, y ≥ 0$.

In reference to the above LPP, which of the following are correct?

(A) The feasible region is bounded.
(B) The corner points of the feasible region are (15, 20), (40, 15) and (0, 75).
(C) The maximum value of the objective function is 285.
(D) The LPP does not have optimal solution.

Choose the correct answer from the options given below:

(A) and (C) only

The correct answer is Option (3) → (A) and (C) only

$\text{Constraints: }4x+y\ge 80,\ x+5y\ge 115,\ 3x+2y\le 150,\ x\ge0,\ y\ge0$

Intersection of pairs (equalities):

$4x+y=80,\ x+5y=115 \Rightarrow (x,y)=(15,20)$

$4x+y=80,\ 3x+2y=150 \Rightarrow (x,y)=(2,72)$

$x+5y=115,\ 3x+2y=150 \Rightarrow (x,y)=(40,15)$

Check feasibility of these points (they satisfy all inequalities):

$(15,20):\ 4(15)+20=80,\ 15+5(20)=115,\ 3(15)+2(20)=85\le150$

$(2,72):\ 4(2)+72=80,\ 2+5(72)=362,\ 3(2)+2(72)=150$

$(40,15):\ 4(40)+15=175,\ 40+5(15)=115,\ 3(40)+2(15)=150$

$\text{Feasible region is the triangle with vertices }(15,20),(2,72),(40,15)\text{, so it is bounded. (A is true)}$

Evaluate $Z=6x+3y$ at these vertices:

$Z(15,20)=6(15)+3(20)=150$

$Z(2,72)=6(2)+3(72)=228$

$Z(40,15)=6(40)+3(15)=285$

Maximum value $=285$ at $(40,15)$, so (C) is true.

Statement (B) is false because $(0,75)$ is not a feasible corner (it fails $4x+y\ge80$).

Statement (D) is false because an optimal solution exists (maximum $285$).

Correct options: (A) and (C)