Question

Find the optimal solution for the following problem

Minimize C = 13x + 3y
subject to 12x + 14y ≥ 21
15x + 20y ≥ 37
and x ≥ 0, y ≥ 0.
1. What is the optimal value of x?

2. What is the optimal value of y?

3.What is the minimum value of the objective function?

Answer 1

Answer:

1. optimal value of x, is 0

2. optimal value of y, is 37/20

3. optimal value of objetive function, is 111/20

Step-by-step explanation:

In the graphic of the attached file you can see the feasible region for the linear programming problem.

The vertices of this region are the origin and the intercepts of the line 1$5x + 20y = 37$. That is, $A (0, 37/20)$ and $B (37/15, 0)$. According to the optimality theorem, the optimal solution of the problem must be reached at one of the vertices of the feasible region, therefore,

We evaluate the objective function $Z = 13x + 3y$ at each vertex:

$Z (0, 37/20) = 13 (0) + 3 (37/20) = 111/20\\\\Z (37/15, 0) = 13 (37/15) + 3 (0) = 481/215$

Since $Z (0, 37/20)$ is less than $Z (37/15, 0)$, then the optimal solution is reached at $x = 0$, $y = 37/20$.

optimal value of x, is 0

optimal value of y, is 37/20

optimal value of objetive function, is 111/20