Question

The minimum value of \( Z = 3x + 5y \), given subject to the constraints \( x + y \geq 2 \), \( x + 3y \geq 3 \), \( x, y \geq 0 \), is:

• A. 6
• B. 8
• C. 9
• D. 7

Hint:

To solve linear programming problems, graph the constraints to find the feasible region, then evaluate the objective function at the vertices of the feasible region. The minimum or maximum value will occur at one of the vertices.

Answer 1

The Correct Option is D

We are given the objective function \( Z = 3x + 5y \) and the following constraints: \[ x + y \geq 2 \] \[ x + 3y \geq 3 \] \[ x, y \geq 0 \] To solve this linear programming problem, we need to find the feasible region and then evaluate the objective function at the vertices of the region. Step 1: Graph the constraints. 1. \( x + y = 2 \) represents a line with intercepts at \( (2, 0) \) and \( (0, 2) \). 2. \( x + 3y = 3 \) represents a line with intercepts at \( (3, 0) \) and \( (0, 1) \). 3. The non-negativity constraints \( x \geq 0 \) and \( y \geq 0 \) restrict the feasible region to the first quadrant. Step 2: Find the intersection points of the lines. 1. Solving \( x + y = 2 \) and \( x + 3y = 3 \) simultaneously: \[ x + y = 2 \quad \text{(A)} \] \[ x + 3y = 3 \quad \text{(B)} \] Subtract equation (A) from equation (B): \[ (x + 3y) - (x + y) = 3 - 2 \] \[ 2y = 1 \quad \Rightarrow \quad y = \frac{1}{2} \] Substitute \( y = \frac{1}{2} \) into equation (A): \[ x + \frac{1}{2} = 2 \quad \Rightarrow \quad x = \frac{3}{2} \] Thus, the intersection point is \( \left( \frac{3}{2}, \frac{1}{2} \right) \). Step 3: Evaluate the objective function at the vertices of the feasible region. The vertices of the feasible region are: - \( (0, 2) \) - \( (3, 0) \) - \( \left( \frac{3}{2}, \frac{1}{2} \right) \) Now, evaluate \( Z = 3x + 5y \) at each vertex: 1. At \( (0, 2) \), \( Z = 3(0) + 5(B) = 10 \) 2. At \( (3, 0) \), \( Z = 3(3) + 5(0) = 9 \) 3. At \( \left( \frac{3}{2}, \frac{1}{2} \right) \), \( Z = 3\left( \frac{3}{2} \right) + 5\left( \frac{1}{2} \right) = \frac{9}{2} + \frac{5}{2} = 7 \) The minimum value of \( Z \) is \( 7 \), which occurs at \( \left( \frac{3}{2}, \frac{1}{2} \right) \). Thus, the correct answer is option (D)