Question

In the graph, the feasible region representing the Linear Programming Problem for maximising objective function \(Z = px + qy, p, q > 0\) is shaded. If all points on segment \(AB\) give max (Z), then which of the following is true ?

• A. \(p = 2q\)
• B. \(p = 3q\)
• C. \(q = 3p\)
• D. \(q = 2p\)

Hint:

When multiple optimal solutions exist on a line segment, the objective function is always parallel to that constraint line.
Check your signs carefully: slopes of lines going "downhill" from left to right must be negative.
If the equation is \(3p = q\), then \(q\) is three times larger than \(p\).

Answer 1

The Correct Option is C

Concept:
• If an objective function \(Z = px + qy\) achieves its maximum value at every point on a line segment, then the objective function must be parallel to that line segment.
• Parallel lines have equal slopes.
• The slope of the objective function \(Z = px + qy\) is \(-\frac{p}{q}\).

Step 1: Identify the coordinates of points A and B from the graph
From the provided graph:
Point \(A\) is the y-intercept of the top boundary line: \(A = (0, 5)\).
Point \(B\) is the intersection of two lines: \(B = (3, 4)\).

Step 2: Calculate the slope of the line segment \(AB\)
The slope \(m\) of the segment joining \((x_1, y_1)\) and \((x_2, y_2)\) is given by:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Substituting coordinates of \(A(0, 5)\) and \(B(3, 4)\):
\[ m_{AB} = \frac{4 - 5}{3 - 0} = \frac{-1}{3} = -\frac{1}{3} \]

Step 3: Find the slope of the objective function
The objective function is \(Z = px + qy\).
To find the slope, rewrite it in \(y = mx + c\) form:
\[ qy = -px + Z \] \[ y = \left(-\frac{p}{q}\right)x + \frac{Z}{q} \] The slope of the objective function is \(-\frac{p}{q}\).

Step 4: Equate the slopes to find the relationship between p and q
Since all points on \(AB\) give the same maximum \(Z\), the slopes must be equal:
\[ -\frac{p}{q} = -\frac{1}{3} \] Cross-multiplying:
\[ 3p = q \] Or \(q = 3p\).
This matches option (C).