Question

Consider the linear programming problem:
Maximize   z=10x+5y
subject to the constraints
2x+3y≤120
2x + y ≤ 60
x,y≥0.
Then the coordinates of the corner points of the feasible region are

• A. (0, 0), (30, 0), (0, 40) and (15, 30)
• B. (0, 0), (60, 0), (0, 40) and (15, 30)
• C. (0, 0), (30, 0), (0, 60) and (15, 30)
• D. (0, 0), (30, 0), (0, 40) and (30, 40)
• E. (0, 0), (60, 0), (0, 40) and (30, 40)

Answer 1

The Correct Option is A

We first graph the constraints to identify the feasible region. The equations for the constraints are:
1. \( 2x + 3y \leq 120 \)
2. \( 2x + y \leq 60 \)
3. \( x, y \geq 0 \)
Now, we find the points of intersection for the corner points of the feasible region.
The first constraint intersects the x-axis at \( y = 0 \), giving the point \( (60, 0) \).
The second constraint intersects the x-axis at \( y = 0 \), giving the point \( (30, 0) \).
The first and second constraints intersect each other by solving the system of equations: \[ 2x + 3y = 120 \quad \text{and} \quad 2x + y = 60 \] Subtract the second equation from the first: \[ 2x + 3y - (2x + y) = 120 - 60 \quad \Rightarrow \quad 2y = 60 \quad \Rightarrow \quad y = 30 \] Substitute \( y = 30 \) into \( 2x + y = 60 \): \[ 2x + 30 = 60 \quad \Rightarrow \quad 2x = 30 \quad \Rightarrow \quad x = 15 \] So, the point of intersection is \( (15, 30) \). Thus, the coordinates of the corner points are \( (0, 0), (30, 0), (0, 40), (15, 30) \).

The correct option is (A) : \((0, 0), (30, 0), (0, 40)\  \text{and} \ (15, 30)\)

Answer 2

We are given a linear programming problem: 

Maximize: \( z = 10x + 5y \)
Subject to:
\( 2x + 3y \leq 120 \)
\( 2x + y \leq 60 \)
\( x \geq 0, y \geq 0 \)

Let's find the feasible region by finding the intersection points (corner points) of the constraints:

Step 1: Find intercepts of the lines
For \( 2x + 3y = 120 \):
- Let \( x = 0 \Rightarrow y = 40 \)
- Let \( y = 0 \Rightarrow x = 60 \)

For \( 2x + y = 60 \):
- Let \( x = 0 \Rightarrow y = 60 \)
- Let \( y = 0 \Rightarrow x = 30 \)

Step 2: Find intersection point of the two lines
Solve: \[ \begin{cases} 2x + 3y = 120 \\ 2x + y = 60 \end{cases} \] Subtract second from first: \[ (2x + 3y) - (2x + y) = 120 - 60 \Rightarrow 2y = 60 \Rightarrow y = 30 \] Substitute in second equation: \[ 2x + 30 = 60 \Rightarrow x = 15 \] So, intersection point is \( (15, 30) \)

Step 3: Corner points of the feasible region
From intercepts and intersection: - \( (0, 0) \) - \( (30, 0) \) - \( (0, 40) \) - \( (15, 30) \)

Correct answer: (0, 0), (30, 0), (0, 40) and (15, 30)