Question

Maximize \( z = x + y \) subject to: \[ x + y \leq 10, \quad 3y - 2x \leq 15, \quad x \leq 6, \quad x, y \geq 0. \] Find the maximum value.

Hint:

When maximizing or minimizing a linear objective function subject to linear constraints, the optimal value occurs at one of the vertices of the feasible region.

Answer 1

Step 1: Understand the constraints. We are given the following constraints for the variables \(x\) and \(y\): \[ x + y \leq 10, \quad 3y - 2x \leq 15, \quad x \leq 6, \quad x, y \geq 0. \] These constraints define the feasible region on a graph. 

Step 2: Identify the vertices of the feasible region. To find the feasible region, plot these constraints on a graph. The vertices where the constraints intersect are \( (0, 0), (0, 5), (6, 4), (6, 0) \). 

Step 3: Evaluate the objective function at the vertices. The objective function is \( z = x + y \). Let's evaluate it at each vertex: 
- At \( (0, 0), z = 0 + 0 = 0 \). 
- At \( (0, 5), z = 0 + 5 = 5 \). 
- At \( (6, 4), z = 6 + 4 = 10 \). 
- At \( (6, 0), z = 6 + 0 = 6 \).


Step 4: Conclusion. The maximum value of \( z \) is 10, which occurs at \( (6, 4) \).