Question

Corner points of a feasible bounded region are \((0, 10)\), \((4, 2)\), \((3, 7)\) and \((10, 6)\). Maximum value 50 of objective function \(z = ax + by\) occurs at two points \((0, 10)\) and \((10, 6)\). The value of \(a\) and \(b\) are:

• A. \(a = 5, b = 2\)
• B. \(a = 4, b = 5\)
• C. \(a = 2, b = 5\)
• D. \(a = 5, b = 4\)

Hint:

For maximum at two points, objective function values are equal at both points.

Answer 1

The Correct Option is A

The objective function is \(z = ax + by\). Given maximum value \(z = 50\) occurs at points \((0, 10)\) and \((10, 6)\). So, \[ z(0,10) = a \cdot 0 + b \cdot 10 = 10b = 50 \implies b = 5 \] and \[ z(10,6) = a \cdot 10 + b \cdot 6 = 10a + 6b = 50 \] Substitute \(b=5\): \[ 10a + 6 \times 5 = 50 \implies 10a + 30 = 50 \implies 10a = 20 \implies a = 2 \] Rechecking the max value at the two points with \(a=2, b=5\): \[ z(0,10) = 0 + 5 \times 10 = 50 \] \[ z(10,6) = 2 \times 10 + 5 \times 6 = 20 + 30 = 50 \] So the values are \(a=2\), \(b=5\). But this matches option (c), not (a). The question states maximum at two points, so the correct pair should satisfy both. So the correct answer is (c) \(a=2, b=5\).