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

Answer: (A)

LPP has a unique optimal solution

Answer 2

Given that:

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. (_Ans.)