Question

Maximize \( Z = 3x + 9y \) subject to the constraints: \[ x + 3y \leq 60, x + y \geq 10, x \leq y, x \geq 0, y \geq 0. \]

Hint:

In linear programming, the optimal solution is found at the corner points of the feasible region.

Answer 1

Step 1: Formulate the problem.
We need to maximize \( Z = 3x + 9y \) under the constraints: - \( x + 3y \leq 60 \), - \( x + y \geq 10 \), - \( x \leq y \), - \( x \geq 0 \), - \( y \geq 0 \).

Step 2: Graph the constraints.
Graph the inequalities on a coordinate plane: - \( x + 3y \leq 60 \) is a line with slope \( -\frac{1}{3} \), - \( x + y \geq 10 \) is a line with slope \( -1 \), - The region of feasible points is where all inequalities are satisfied, which is the region on or below these lines.

Step 3: Identify the corner points.
The corner points of the feasible region are the points of intersection of the lines: - \( x + 3y = 60 \) intersects \( x + y = 10 \) at \( (30, 10) \), - The intercepts for the lines give additional points at \( (0, 10) \) and \( (60, 0) \).

Step 4: Calculate the value of \( Z \) at each corner point.
At \( (30, 10) \), \( Z = 3(30) + 9(10) = 90 \), At \( (0, 10) \), \( Z = 3(0) + 9(10) = 90 \), At \( (60, 0) \), \( Z = 3(60) + 9(0) = 180 \).

Step 5: Conclusion.
The maximum value of \( Z \) is \( 180 \), which occurs at the point \( (60, 0) \).