Question

Solve the following L.P.P. by graphical method: 
Maximize: 
\[ z = 10x + 25y. \] Subject to: \[ 0 \leq x \leq 3, \quad 0 \leq y \leq 3, \quad x + y \leq 5. \]

Hint:

For L.P.P, the optimal solution is found by evaluating the objective function at the vertices of the feasible region. The maximum (or minimum) value occurs at one of the corner points of the feasible region.

Answer 1

To determine the maximum value of \(Z\), we evaluate it at the vertices of the feasible region.

Objective function:

\[ Z = 10x + 25y \]

Evaluating \(Z\) at the given points:

\[ Z \text{ at } O(0,0) = 10(0) + 25(0) = 0 \] \[ Z \text{ at } A(3,0) = 10(3) + 25(0) = 30 \] \[ Z \text{ at } B(3,2) = 10(3) + 25(2) = 30 + 50 = 80 \] \[ Z \text{ at } C(2,3) = 10(2) + 25(3) = 20 + 75 = 95 \] \[ Z \text{ at } D(0,3) = 10(0) + 25(3) = 75 \] Conclusion:

The maximum value of \(Z\) is 95 at \(C(2,3)\).

Thus, \(Z\) is maximum when \(x = 2\) and \(y = 3\).