Solution: The slope of the objective function \( Z = \alpha x + \beta y \) is given by \( -\frac{\alpha}{\beta} \). To maximize \( Z \), the slope of the objective function must match the slope of the line passing through the points (5, 5) and (0, 20).
The slope of the line passing through (5, 5) and (0, 20) is:
\(\text{Slope} = \frac{20 - 5}{0 - 5} = -3\).
Equating this with the slope of the objective function:
\(-\frac{\alpha}{\beta} = -3 \implies \alpha = 3\beta\).
Thus, the correct answer is \( \alpha = 3\beta \).
Minimize Z = 5x + 3y \text{ subject to the constraints} \[ 4x + y \geq 80, \quad x + 5y \geq 115, \quad 3x + 2y \leq 150, \quad x \geq 0, \quad y \geq 0. \]
Solve the following L.P.P. by graphical method:
Maximize:
\[ z = 10x + 25y. \] Subject to: \[ 0 \leq x \leq 3, \quad 0 \leq y \leq 3, \quad x + y \leq 5. \]
Find the minimum value of ( z = x + 3y ) under the following constraints:
• x + y ≤ 8
• 3x + 5y ≥ 15
• x ≥ 0, y ≥ 0