Question:

In an LPP, corner points of the feasible region determined by the system of linear constraints are \((1,1), (3,0), (0,3)\). If \(Z = ax + by\), where \(a>0, b>0\) is to be minimized, the condition on \(a\) and \(b\) so that the minimum of \(Z\) occurs at \((3, 0)\) and \((1, 1)\) will be:

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In linear programming problems, if the minimum or maximum occurs at more than one point, set the values of the objective function equal at those points to find the required condition.
  • \(a = 2b\)
  • \(a = \dfrac{b}{2}\)
  • \(a = 3b\)
  • \(a = b\)
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The Correct Option is B

Solution and Explanation

We are given the objective function: \[ Z = ax + by \] and it is to be minimized over the feasible region formed by the points: \[ A(1,1), \quad B(3,0), \quad C(0,3) \] Let us evaluate \(Z = ax + by\) at each of the corner points: At point \(A(1,1)\): \[ Z_1 = a(1) + b(1) = a + b \] At point \(B(3,0)\): \[ Z_2 = a(3) + b(0) = 3a \] At point \(C(0,3)\): \[ Z_3 = a(0) + b(3) = 3b \] We are told that the minimum of \(Z\) occurs at both \((3,0)\) and \((1,1)\). So we must have: \[ Z_1 = Z_2 \Rightarrow a + b = 3a \Rightarrow b = 2a \Rightarrow a = \frac{b}{2} \] \[ \boxed{a = \frac{b}{2}} \]
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