The maximum value of \(Z = 4x + 3y\) subject to the constraints \(x + 2y \le k\), \(x \le 6\), \(x \ge 0\), \(y \ge 0\), where \(k \gt 6\), is 30. Then the value of k is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 10
Since \(k \gt 6\), the line \(x = 6\) meets \(x + 2y = k\) at \(\left(6, \frac{k - 6}{2}\right)\). The corner points are (0, 0), (6, 0), \(\left(6, \frac{k - 6}{2}\right)\) and \(\left(0, \frac{k}{2}\right)\). Values of Z:
\(Z(0, 0) = 0\)
\(Z(6, 0) = 24\)
\(Z\left(6, \frac{k - 6}{2}\right) = 24 + \frac{3k - 18}{2} = \frac{3k + 30}{2}\)
\(Z\left(0, \frac{k}{2}\right) = \frac{3k}{2}\)
Since \(k \gt 6\), \(\frac{3k + 30}{2}\) is larger than both 24 and \(\frac{3k}{2}\), so it is the maximum:
\(\frac{3k + 30}{2} = 30\)
\(3k = 30\)
\(k = 10\)
Check: the corner is then (6, 2), and \(Z = 24 + 6 = 30\).
Why the other options are incorrect
Option (1) 8 — Incorrect. The maximum would be \(\frac{24 + 30}{2} = 27\).
Option (3) 12 — Incorrect. The maximum would be \(\frac{36 + 30}{2} = 33\).
Option (4) 20 — Incorrect. This comes from \(\frac{3k}{2} = 30\), the value at \(\left(0, \frac{k}{2}\right)\), which is not the maximum; for \(k = 20\) the maximum is 45.