Question

Minimize \( Z = 3x + 5y \) subject to the constraints: \[ x + 3y \geq 3, x + y \geq 2, x \geq 0, y \geq 0. \]

Hint:

In linear programming problems, the optimal solution lies at the corner points of the feasible region.

Answer 1

Step 1: Formulate the problem.
We need to minimize \( Z = 3x + 5y \) under the constraints: - \( x + 3y \geq 3 \), - \( x + y \geq 2 \), - \( x \geq 0 \), - \( y \geq 0 \).

Step 2: Graph the constraints.
Graph the inequalities on a coordinate plane: - \( x + 3y \geq 3 \) is a line with slope \( -\frac{1}{3} \), - \( x + y \geq 2 \) is a line with slope \( -1 \), - The region of feasible points is where all inequalities are satisfied, which is the region that lies on or above these lines.

Step 3: Identify the corner points.
The corner points of the feasible region are the points of intersection of the lines: - \( x + 3y = 3 \) intersects \( x + y = 2 \) at \( (x, y) = (1, 1) \), - The intercepts for the lines give additional points at \( (3, 0) \) and \( (0, 1) \).

Step 4: Calculate the value of \( Z \) at each corner point.
At \( (1, 1) \), \( Z = 3(1) + 5(1) = 8 \), At \( (3, 0) \), \( Z = 3(3) + 5(0) = 9 \), At \( (0, 1) \), \( Z = 3(0) + 5(1) = 5 \).

Step 5: Conclusion.
The minimum value of \( Z \) is \( 5 \), which occurs at the point \( (0, 1) \).