91Ó°ÊÓ

Since we have constraints and we need to identify the area they define. We'll first visualize the constraints on a 3D coordinate system, focusing on the intersection of the planes formed by the first two constraints and their relationship with the non-negativity constraints.

The feasible region is the intersection of the constraints, where all constraints hold true: 1. x + y + z = 100 (since x, y, and z are all non-negative, and we wish to find the largest value of p) 2. x + y + z ≤ 150 (we need to find the corner points where the equation equals 150) 3. x, y, and z are non-negative To find the corner points of the feasible region, we can consider the edges of the feasible region where at least two of x, y, or z are equal to 0: 1. x = 0 and y = 0: (0, 0, 100) 2. x = 0 and z = 0: (0, 100, 0) 3. y = 0 and z = 0: (100, 0, 0) 4. x = 0 and y + z = 150: (0, 150, 0) 5. x = 0 and x + z = 150: (0, 0, 150) 6. y = 0 and x + z = 150: (150, 0, 0) 7. y = 0 and x + y = 150: (150, 0, 0) Note that point (0, 150, 0) satisfies both constraint (3) and constraint (4) as it lies on the boundary of constraint (2) and satisfies constraint (1). Corner points are (0, 0, 100), (0, 100, 0), (100, 0, 0), and (0, 150, 0).

Now we will evaluate the objective function, p = 2x + 5y + 3z, at each of the identified corner points: 1. p(0, 0, 100) = 2(0) + 5(0) + 3(100) = 300 2. p(0, 100, 0) = 2(0) + 5(100) + 3(0) = 500 3. p(100, 0, 0) = 2(100) + 5(0) + 3(0) = 200 4. p(0, 150, 0) = 2(0) + 5(150) + 3(0) = 750

Comparing the values of the objective function at each corner point, we find that the maximum value for p is 750, occurring at point (0, 150, 0). Thus, the optimal solution for this linear programming problem is x=0, y=150, and z=0, which results in a maximum value of p = 750.