The corner points of the feasible region for an L.P.P. are (0, 10), (5, 5), (5, 15) and (0, 30). If the objective function is Z = αx + βy, α, β > 0, the condition on α and β so that the maximum of Z occurs at corner points (5, 5) and (0, 20) is :
α = 3β
This problem involves a Linear Programming Problem (LPP) where we are given some corner points of the feasible region and an objective function $Z = \alpha x + \beta y$, with $\alpha > 0$ and $\beta > 0$. We are told that the maximum value of this objective function Z occurs at two specific points, (5, 5) and (0, 20). We need to find the relationship between $\alpha$ and $\beta$ based on this condition.
The corner points of the feasible region are given as (0, 10), (5, 5), (5, 15), and (0, 30). However, the question states that the maximum of Z occurs at corner points (5, 5) and (0, 20). In LPP, if the maximum occurs at two distinct points, it means the objective function takes the same maximum value at both points.
In a Linear Programming Problem, if the optimal solution (maximum or minimum) occurs at two distinct points on the boundary of the feasible region, then it occurs at every point on the line segment joining these two points. This happens when the line representing the objective function for the optimal value is parallel to a boundary edge of the feasible region, and this edge contains the two points.
A fundamental property in this case is that the value of the objective function is identical at these two distinct optimal points.
The objective function is $Z = \alpha x + \beta y$. We are given that the maximum value of Z occurs at points (5, 5) and (0, 20).
Let's evaluate the objective function Z at each of these points:
Since the maximum value of Z occurs at both (5, 5) and (0, 20), the value of the objective function must be the same at these two points. We set the expressions for $Z(5, 5)$ and $Z(0, 20)$ equal:
$\qquad 5\alpha + 5\beta = 20\beta$
Now we solve this equation to find the required relationship between $\alpha$ and $\beta$.
Subtract $5\beta$ from both sides of the equation:
$\qquad 5\alpha = 20\beta - 5\beta$
$\qquad 5\alpha = 15\beta$
To isolate $\alpha$, divide both sides of the equation by 5:
$\qquad \alpha = \frac{15\beta}{5}$
$\qquad \alpha = 3\beta$
This equation $\alpha = 3\beta$ represents the condition on $\alpha$ and $\beta$ for the maximum of the objective function $Z = \alpha x + \beta y$ to occur at both points (5, 5) and (0, 20).
Let's compare the derived condition $\alpha = 3\beta$ with the provided options:
The condition we found, $\alpha = 3\beta$, matches Option 3.
This condition implies that the ratio $\alpha/\beta = 3$. Since $\alpha, \beta > 0$, the slope of the objective function line $Z = \alpha x + \beta y$ (which is $y = -\frac{\alpha}{\beta}x + \frac{Z}{\beta}$, with slope $-\frac{\alpha}{\beta}$) is $-\frac{3\beta}{\beta} = -3$. The slope of the line segment connecting (5, 5) and (0, 20) is $\frac{20-5}{0-5} = \frac{15}{-5} = -3$. The equality of slopes confirms that the objective function line is parallel to the segment connecting these two points, consistent with the maximum occurring along this segment.
| LPP Term | Explanation |
|---|---|
| Objective Function | The function to maximize or minimize (e.g., $Z = ax + by$). |
| Constraints | Inequalities or equations that define the limitations on the variables. |
| Feasible Region | The graphical area satisfying all constraints. It's a convex set. |
| Corner Points | Vertices of the feasible region polygon. Optimal solutions lie here. |
| Optimal Solution | The point(s) in the feasible region where the objective function reaches its maximum or minimum value. |
| Multiple Optima | When the objective function attains the optimal value at more than one point, typically along a boundary edge. |
Linear Programming is a powerful mathematical technique used across various fields, including business, economics, and engineering, for decision-making involving optimization.
The problem demonstrates a key characteristic of LPP optima occurring along an edge when the objective function's slope aligns with that edge.
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