Question

Consider the following Linear Programming Problem(LPP):Maximize \(z=60x_1+50x_2\) subject to \(x_1+2x_2≤40 3x_1+2x_2≤60 x_1,x_2≥0\).Then, the ______.

• A. LPP has a unique optimal solution
• B. LPP is infeasible.
• C. LPP is unbounded
• D. LPP has multiple optimal solutions
• E. LPP has no solution

Answer 1

The Correct Option is A

Given: The given LPP is as follows: Maximize \(z = 60x_1 + 50x_2\) 

Subject to the constraints: x_1 + 2x_2 ≤ 40  , 3x_1 + 2x_2 ≤ 60 ,   x_1, x_2 ≥ 0

We can graph the feasible region defined by these constraints in the\(x_1, x_2\) plane to visualize the problem:

Plot the lines \(x_1 + 2x_2 = 40\) and \(3x_1 + 2x_2 = 60\).

Then,

Shade the region below both lines (since they are inequalities, the feasible region is below the lines).

Then,

The feasible region is bounded by the \(x_1\) and \(x_2\) axes, as both variables are non-negative \((x_1, x_2 ≥ 0)\).

The feasible region will look like a triangular area in the first quadrant.

Now, we need to find the optimal solution. To do that, we evaluate the objective function (z = 60x_1 + 50x_2) at each corner point (vertex) of the feasible region, as there are only a finite number of corner points.

Corner points of the feasible region:

\((0, 0)\)- The origin

\((0, 20)\) - The intersection of \(x_1\)-axis and the first constraint

\((20, 0)\) - The intersection of \(x_2\)-axis and the first constraint

\((10, 15)\)- The intersection of the two constraints

Now, we calculate z for each corner point:

\(z(0, 0) = 60 × 0 + 50  0 = 0\)

\(z(0, 20) = 60 × 0 + 50 × 20 = 1000\)

\(z(20, 0) = 60 × 20 + 50 × 0 = 1200\)

\(z(10, 15) = 60 × 10 + 50 × 15 = 1350\)

The max. value of \(z\) occurs at point \((10, 15)\), where \(z = 1350.\)

Now, since the objective function has a unique maximum value at a specific point, and the feasible region is bounded (as shown by the graph), we can conclude that the LPP has a unique optimal solution.

Answer 2

Let's analyze the given Linear Programming Problem (LPP):

Maximize \( z = 60x_1 + 50x_2 \)

Subject to:

\[ x_1 + 2x_2 \leq 40 \] \[ 3x_1 + 2x_2 \leq 60 \] \[ x_1, x_2 \geq 0 \]

We can solve this graphically. First, let's find the feasible region by plotting the constraints:

1. \( x_1 + 2x_2 \leq 40 \): This line passes through \( (40, 0) \) and \( (0, 20) \). The feasible region is below this line.
2. \( 3x_1 + 2x_2 \leq 60 \): This line passes through \( (20, 0) \) and \( (0, 30) \). The feasible region is below this line.
3. \( x_1 \geq 0 \) and \( x_2 \geq 0 \): This restricts the feasible region to the first quadrant.

The feasible region is a quadrilateral with vertices at \( (0, 0) \), \( (20, 0) \), \( (0, 20) \), and the intersection point of \( x_1 + 2x_2 = 40 \) and \( 3x_1 + 2x_2 = 60 \). 

Subtracting the first equation from the second gives \( 2x_1 = 20 \), so \( x_1 = 10 \). Substituting into the first equation gives \( 10 + 2x_2 = 40 \), so \( 2x_2 = 30 \), and \( x_2 = 15 \). 

Thus, the intersection point is \( (10, 15) \).

Now we evaluate the objective function \( z \) at each vertex:

• \( (0, 0) \): \( z = 0 \)
• \( (20, 0) \): \( z = 60(20) + 50(0) = 1200 \)
• \( (0, 20) \): \( z = 60(0) + 50(20) = 1000 \)
• \( (10, 15) \): \( z = 60(10) + 50(15) = 600 + 750 = 1350 \)

The maximum value of \( z \) is 1350 at the point \( (10, 15) \). This is a unique optimal solution.

Therefore, the correct statement is: The LPP has a unique optimal solution.