Consider the linear programming problem(LPP):

Minimize $Z = x + y$

$x + 2y \leq 4,$

$3x + y \geq 3,$

$4x + 3y \geq 6,$

$x, y \geq 0.$

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:

We need to minimize $Z = x + y$ subject to:

$x + 2y \leq 4, \quad 3x + y \geq 3, \quad 4x + 3y \geq 6, \quad x, y \geq 0$

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Intersection of $4x + 3y = 6$ and $y = 0$:

$4x = 6$

$x = \tfrac{3}{2}$

Corner point: $\left(\tfrac{3}{2},\; 0\right)$

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Intersection of $4x + 3y = 6$ and $3x + y = 3$:

From $3x + y = 3$: $\;y = 3 - 3x$

$4x + 3(3 - 3x) = 6$

$4x + 9 - 9x = 6$

$-5x = -3$

$x = \tfrac{3}{5}$

$y = 3 - \tfrac{9}{5} = \tfrac{6}{5}$

Corner point: $\left(\tfrac{3}{5},\; \tfrac{6}{5}\right)$

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Intersection of $3x + y = 3$ and $x + 2y = 4$:

From $3x + y = 3$: $\;y = 3 - 3x$

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

$x + 6 - 6x = 4$

$-5x = -2$

$x = \tfrac{2}{5}$

$y = 3 - \tfrac{6}{5} = \tfrac{9}{5}$

Corner point: $\left(\tfrac{2}{5},\; \tfrac{9}{5}\right)$

Note: this point is $\left(\tfrac{2}{5},\;\tfrac{9}{5}\right)$, not $\left(\tfrac{2}{5},\;\tfrac{6}{5}\right)$.

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Intersection of $x + 2y = 4$ and $y = 0$:

$x = 4$

Corner point: $\left(4,\; 0\right)$

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The corner points are: $\left(\tfrac{3}{2}, 0\right)$, $\left(\tfrac{3}{5}, \tfrac{6}{5}\right)$, $\left(\tfrac{2}{5}, \tfrac{9}{5}\right)$, and $(4, 0)$.

Since all four corners close off the region, the feasible region is bounded.

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

At $\left(\tfrac{3}{2},\; 0\right)$: $Z = \tfrac{3}{2} = 1.5$

At $\left(\tfrac{3}{5},\; \tfrac{6}{5}\right)$: $Z = \tfrac{9}{5} = 1.8$

At $\left(\tfrac{2}{5},\; \tfrac{9}{5}\right)$: $Z = \tfrac{11}{5} = 2.2$

At $(4,\; 0)$: $Z = 4$

$Z_{\min} = \dfrac{3}{2}$ at $\left(\dfrac{3}{2},\; 0\right)$

Since no other corner gives the same minimum and $Z = x + y = \tfrac{3}{2}$ is not parallel to any edge touching this corner, the solution is unique.

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(A) Bounded feasible region — Correct, the region is enclosed on all sides.

(B) Unique optimal solution — Correct, only one point gives $Z = \tfrac{3}{2}$.

(C) Optimal at $\left(\tfrac{3}{2}, 0\right)$ — Correct.

(D) Corner points include $\left(\tfrac{2}{5}, \tfrac{6}{5}\right)$ — Wrong, the correct point is $\left(\tfrac{2}{5}, \tfrac{9}{5}\right)$.

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The answer is (A), (B) and (C) only.