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:

The constraints and their boundary lines are:

$4x + y \geq 80 \implies L_1: 4x + y = 80$

$x + 5y \geq 115 \implies L_2: x + 5y = 115$

$3x + 2y \leq 150 \implies L_3: 3x + 2y = 150$

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Finding intersection of $L_1$ and $L_2$:

From $L_1$: $y = 80 - 4x$

$x + 5(80 - 4x) = 115$

$x + 400 - 20x = 115$

$-19x = -285$

$x = 15$

$y = 80 - 4(15) = 20$

Point: $(15, 20)$

Check $L_3$: $3(15) + 2(20) = 85 \leq 150$ ✔️

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Finding intersection of $L_1$ and $L_3$:

From $L_1$: $y = 80 - 4x$

$3x + 2(80 - 4x) = 150$

$3x + 160 - 8x = 150$

$-5x = -10$

$x = 2$

$y = 80 - 4(2) = 72$

Point: $(2, 72)$

Check $L_2$: $2 + 5(72) = 362 \geq 115$ ✔️

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Finding intersection of $L_2$ and $L_3$:

From $L_2$: $x = 115 - 5y$

$3(115 - 5y) + 2y = 150$

$345 - 15y + 2y = 150$

$-13y = -195$

$y = 15$

$x = 115 - 5(15) = 40$

Point: $(40, 15)$

Check $L_1$: $4(40) + 15 = 175 \geq 80$ ✔️

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Checking axis intersection points for feasibility:

$(0, 80)$ from $L_1$: $3(0) + 2(80) = 160 > 150$ ❌

$(0, 75)$ from $L_3$: $4(0) + 75 = 75 < 80$ ❌

$(20, 0)$ from $L_1$: $20 + 0 = 20 < 115$ ❌

$(50, 0)$ from $L_3$: $50 + 0 = 50 < 115$ ❌

None of the axis points are feasible.

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The three corner points of the feasible region are:

$(15, 20), \quad (2, 72), \quad (40, 15)$

All three bounding lines enclose a triangle with no open direction, so the region is bounded.

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Evaluating $Z = 6x + 3y$ at each corner:

At $(15, 20)$:

$Z = 6(15) + 3(20) = 90 + 60 = 150$

At $(2, 72)$:

$Z = 6(2) + 3(72) = 12 + 216 = 228$

At $(40, 15)$:

$Z = 6(40) + 3(15) = 240 + 45 = 285$

$Z_{\max} = 285$ at $(40, 15)$

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(A) Feasible region is bounded — Correct, it is a closed triangle.

(B) Corners are $(15,20)$, $(40,15)$, $(0,75)$ — Wrong, $(0, 75)$ is not feasible since $4(0) + 75 = 75 < 80$. The correct third corner is $(2, 72)$.

(C) Maximum value of $Z$ is $285$ — Correct.

(D) LPP has no optimal solution — Wrong, optimal solution exists at $(40, 15)$ with $Z = 285$.

Therefore, (A) and (C) only are correct.