The maximum value of \(Z = px + 2y\) \((p \gt 0)\), subject to the constraints \(x + y \le 5\), \(x \ge 0\), \(y \ge 0\), is 15. Then the value of p is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 3
The feasible region is the triangle with corner points (0, 0), (5, 0) and (0, 5).
\(Z(0, 0) = 0\)
\(Z(5, 0) = 5p\)
\(Z(0, 5) = 10\)
The maximum is the larger of 5p and 10. Since 10 ≠ 15, the maximum 15 must come from the corner (5, 0):
\(5p = 15\)
\(p = 3\)
Check: with p = 3, the corner values are 0, 15 and 10, so the maximum is 15.
Why the other options are incorrect
Option (1) 1 — Incorrect. Then Z(5, 0) = 5 and Z(0, 5) = 10, so the maximum is 10, not 15.
Option (2) 2 — Incorrect. Then Z(5, 0) = 10 and Z(0, 5) = 10, so the maximum is 10.
Option (3) 5 — Incorrect. Then Z(5, 0) = 25, so the maximum is 25.