Question

The maximum value of \(Z = 3x + 4y\) subject to the constraints \(x + y \leq 1\), \(x \geq 0\), \(y \geq 0\) is:

• A. 3
• B. 4
• C. 7
• D. 0

Hint:

In Linear Programming, always evaluate the objective function at the vertices of the feasible region. The maximum or minimum will occur at one of these points.

Answer 1

The Correct Option is B

We are given the objective function: \[ Z = 3x + 4y \] Subject to the constraints: \[ x + y \leq 1, \quad x \geq 0, \quad y \geq 0 \] These constraints define a feasible region in the first quadrant bounded by the line \(x + y = 1\), the \(x\)-axis and the \(y\)-axis. Let us identify the corner points (vertices) of the feasible region: 1. When \(x = 0\): \[ x + y = 1 \Rightarrow y = 1 \Rightarrow (0, 1) \] 2. When \(y = 0\): \[ x + y = 1 \Rightarrow x = 1 \Rightarrow (1, 0) \] 3. Intersection of \(x = 0\) and \(y = 0\): \((0, 0)\) So, the corner points are: \[ (0,0),\ (1,0),\ (0,1) \] Now, evaluate \(Z = 3x + 4y\) at each vertex: - At \((0,0)\): \(Z = 3(0) + 4(0) = 0\) - At \((1,0)\): \(Z = 3(1) + 4(0) = 3\) - At \((0,1)\): \(Z = 3(0) + 4(1) = 4\) Maximum value of \(Z\) is: \[ \boxed{4 \text{ at } (0,1)} \]