Consider the linear programming problem with objective function \(Z = 3x + 5y\) and constraints \(x + y \ge 6\), \(x + 3y \ge 10\), \(x \ge 0\), \(y \ge 0\). Consider the following statements:
(A) The feasible region is unbounded.
(B) The minimum value of Z is 22.
(C) The maximum value of Z is 30.
(D) The minimum value of Z occurs at \((10, 0)\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A) and (B) only
Both constraints are of the “\(\ge\)” type, so the region lies above both lines and extends without limit: it is unbounded.
The lines \(x + y = 6\) and \(x + 3y = 10\) meet where
\(2y = 4\)
\(y = 2, \; x = 4\)
The corner points are (0, 6), (4, 2) and (10, 0). Values of Z:
\(Z(0, 6) = 30\)
\(Z(4, 2) = 12 + 10 = 22\)
\(Z(10, 0) = 30\)
Since the region is unbounded, 22 is the minimum only if the half-plane \(3x + 5y \lt 22\) has no point in common with the region. Every feasible point satisfies
\(3x + 5y = 2(x + y) + (x + 3y) \ge 12 + 10 = 22\)
So the minimum is 22. Z has no maximum, because at (x, 0) with \(x \ge 10\), \(Z = 3x\) grows without limit.
So the region is unbounded, the minimum of Z is 22 at (4, 2), and Z has no maximum value.
Checking the statements
(A) The feasible region is unbounded. — Correct. It extends without limit beyond both lines.
(B) The minimum value of Z is 22. — Correct. \(Z(4, 2) = 22\), and \(3x + 5y \ge 22\) at every feasible point.
(C) The maximum value of Z is 30. — Incorrect. At (20, 0), \(Z = 60\); Z has no maximum.
(D) The minimum value of Z occurs at \((10, 0)\). — Incorrect. \(Z(10, 0) = 30 \gt 22\).