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:
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For maximum at two points, objective function values are equal at both points.
The objective function is \(z = ax + by\). The maximum value \(z = 50\) occurs at points \((0, 10)\) and \((10, 6)\). Applying the first point: \[z(0,10) = a \cdot 0 + b \cdot 10 = 10b = 50 \implies b = 5\]Applying the second point: \[z(10,6) = a \cdot 10 + b \cdot 6 = 10a + 6b = 50\]Substituting \(b=5\) into the second equation:\[10a + 6 \times 5 = 50 \implies 10a + 30 = 50 \implies 10a = 20 \implies a = 2\]Verifying the maximum value with \(a=2, b=5\) at the given points:\[z(0,10) = 0 + 5 \times 10 = 50\]\[z(10,6) = 2 \times 10 + 5 \times 6 = 20 + 30 = 50\]Thus, \(a=2\) and \(b=5\). This solution corresponds to option (c). Since the problem specifies the maximum occurs at two points, the correct pair of values for \(a\) and \(b\) must satisfy both conditions. Therefore, the correct answer is (c) \(a=2, b=5\).