Question

Given \( Z = 80x + 120y \), subject to constraints are \( x + 3y \leq 30; 3x + 4y \leq 60; x \geq 0; y \geq 0 \). \( P \) is one of the corner points of the feasible region for the given Linear Programming Problem. Then the coordinate of \( P \) is

• A. \( (0,15) \)
• B. \( (20,0) \)
• C. \( (6,12) \)
• D. \( (30,0) \)

Hint:

In Linear Programming, corner points are obtained from intercepts and intersections of boundary linesAlways check each point in all constraints before selecting the feasible corner point.

Answer 1

The Correct Option is B

Step 1: Write the given constraints.
\[ x + 3y \leq 30 \]
\[ 3x + 4y \leq 60 \]
\[ x \geq 0,\quad y \geq 0 \]

Step 2: Find intercepts of the first constraint.
For \( x + 3y = 30 \):
When \( x = 0 \):
\[ 3y = 30 \]
\[ y = 10 \]
So one intercept is:
\[ (0,10) \]
When \( y = 0 \):
\[ x = 30 \]
So another intercept is:
\[ (30,0) \]

Step 3: Find intercepts of the second constraint.
For \( 3x + 4y = 60 \):
When \( x = 0 \):
\[ 4y = 60 \]
\[ y = 15 \]
So one intercept is:
\[ (0,15) \]
When \( y = 0 \):
\[ 3x = 60 \]
\[ x = 20 \]
So another intercept is:
\[ (20,0) \]

Step 4: Check the feasible point on the \( x \)-axis.
On the \( x \)-axis, \( y = 0 \).
The constraints become:
\[ x \leq 30 \]
and
\[ 3x \leq 60 \]
\[ x \leq 20 \]
Therefore, the maximum feasible value on the \( x \)-axis is:
\[ x = 20 \]
So the corner point on the \( x \)-axis is:
\[ (20,0) \]

Step 5: Check other given options quickly.
For \( (0,15) \):
\[ x + 3y = 0 + 45 = 45 > 30 \]
So \( (0,15) \) is not feasible.
For \( (30,0) \):
\[ 3x + 4y = 90 + 0 = 90 > 60 \]
So \( (30,0) \) is not feasible.

Step 6: Identify the required corner point \( P \).
Since \( (20,0) \) satisfies all constraints and lies on the boundary of the feasible region, it is a corner point.
\[ P = (20,0) \]

Step 7: Final conclusion.
\[ \boxed{(20,0)} \]