Solve the following linear programming problem graphically:
Maximize \( z = x + y \), subject to constraints:
\[ 2x + 5y \leq 100, \quad 8x + 5y \leq 200, \quad x \geq 0, \quad y \geq 0. \]
Step 1: Plot the constraints
The given constraints are: \( 2x + 5y = 100 \): The line passes through \( (50, 0) \) and \( (0, 20) \).
\( 8x + 5y = 200 \): The line passes through \( (25, 0) \) and \( (0, 40) \).
\( x \geq 0 \) and \( y \geq 0 \): Restricts the feasible region to the first quadrant.
The feasible region is the intersection of all these constraints, forming a polygon bounded by the vertices.
Step 2: Find corner points
The corner points of the feasible region are determined by solving the equations pairwise: \( O(0, 0) \): Intersection of \( x = 0 \) and \( y = 0 \).
\( A(25, 0) \): Intersection of \( 8x + 5y = 200 \) and \( y = 0 \).
\( B\left(\frac{50}{3}, \frac{40}{3}\right) \): Intersection of \( 2x + 5y = 100 \) and \( 8x + 5y = 200 \).
\( C(0, 20) \): Intersection of \( 2x + 5y = 100 \) and \( x = 0 \).
Step 3: Evaluate the objective function at corner points
Substitute the coordinates of the vertices into \( z = x + y \): \[ \begin{array}{|c|c|} \hline \text{Corner Point} & \text{Value of } z = x + y \\ \hline O(0, 0) & 0 + 0 = 0 \\ A(25, 0) & 25 + 0 = 25 \\ B\left(\frac{50}{3}, \frac{40}{3}\right) & \frac{50}{3} + \frac{40}{3} = \frac{90}{3} = 30 \\ C(0, 20) & 0 + 20 = 20 \\ \hline \end{array} \]
Step 4: Find the maximum value of \( z \)
The maximum value of \( z \) is \( 30 \), which occurs at \( B\left(\frac{50}{3}, \frac{40}{3}\right) \).
Conclusion: The maximum value of \( z \) is \( 30 \) when \( x = \frac{50}{3} \) and \( y = \frac{40}{3} \).