Consider the feasible region determined by the constraints \(x + 4y \ge 12\), \(2x + y \ge 10\), \(x \ge 0\), \(y \ge 0\), and the following statements:
(A) The minimum value of \(Z = 2x + 3y\) on this region is 14.
(B) The minimum value of \(Z = x + y\) on this region is 6.
(C) \(Z = x - y\) has a minimum value on this region.
(D) The maximum value of \(Z = -x - 4y\) on this region is \(-12\).
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (D) only
The lines \(x + 4y = 12\) and \(2x + y = 10\) meet where
\(2(12 - 4y) + y = 10\)
\(7y = 14\)
\(y = 2, \; x = 4\)
The region is unbounded, with corner points (0, 10), (4, 2) and (12, 0). On an unbounded region, the best corner value is the optimum only if no feasible point does better; this is tested with the two constraints.
For \(2x + 3y\): corner values 30, 14, 24, and
\(7(2x + 3y) = 4(x + 4y) + 5(2x + y) \ge 48 + 50 = 98\)
For \(x + y\): corner values 10, 6, 12, and
\(7(x + y) = (x + 4y) + 3(2x + y) \ge 12 + 30 = 42\)
For \(-x - 4y\): directly from the first constraint,
\(-x - 4y \le -12\), with equality at (4, 2) and (12, 0)
For \(x - y\): at the feasible points (0, y) with \(y \ge 10\), \(Z = -y\), which decreases without limit.
So \(2x + 3y\) has minimum 14, \(x + y\) has minimum 6, \(-x - 4y\) has maximum \(-12\), and \(x - y\) has no minimum.
Checking the statements
(A) The minimum value of \(Z = 2x + 3y\) on this region is 14. — Correct. \(Z(4, 2) = 14\), and \(2x + 3y \ge 14\) at every feasible point.
(B) The minimum value of \(Z = x + y\) on this region is 6. — Correct. \(Z(4, 2) = 6\), and \(x + y \ge 6\) at every feasible point.
(C) \(Z = x - y\) has a minimum value on this region. — Incorrect. At (0, 100), \(Z = -100\); Z can be made as small as we like.
(D) The maximum value of \(Z = -x - 4y\) on this region is \(-12\). — Correct. \(x + 4y \ge 12\) gives \(Z \le -12\), and \(Z(12, 0) = -12\).