again, do not understand how to complete this type of problem Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. HINT [See Example 1.] (Enter EMPTY if the region is empty. Enter UNBOUNDED if the function is unbounded.)Minimize c = x + y subject tox + 2y 2 32x + y2 3x2 0, y 2 0.(x,y) - (

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Description: again, do not understand how to complete this type of problem Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. HINT [See Example 1.] (Enter EMPTY if the region

Answered: Solve the LP problem. If no optimal…

College Algebra

7th Edition

ISBN:9781305115545

Author:James Stewart, Lothar Redlin, Saleem Watson

Publisher:James Stewart, Lothar Redlin, Saleem Watson

Chapter5: Systems Of Equations And Inequalities

Section: Chapter Questions

Problem 4P

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again, do not understand how to complete this type of problem

Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. HINT [See Example 1.] (Enter EMPTY if the region is empty. Enter UNBOUNDED if the function is unbounded.)
Minimize c = x + y subject to
x + 2y 2 3
2x + y2 3
x2 0, y 2 0.
(x,y) - (

Expert Solution

Step 1

The given linear programming problem is as follows.

Minimize $c = x + y$

Subject to the constraints

$x + 2 y \geq 3$

$2 x + y \geq 3$

$x \geq 0$

$y \geq 0$

The solution of the constraint $x + 2 y \geq 3$ consists of all points on the line $x + 2 y = 3$ and the half-plane determined by this line satisfying the inequality $x + 2 y > 3$ .

Consider the equation $x + 2 y = 3$ corresponding to the constraint $x + 2 y \geq 3$ .

When the value $x = 3$ , we have $3 + 2 y = 3 \Rightarrow y = 0$ .

When the value $x = 1$ , we have $1 + 2 y = 3 \Rightarrow y = 1$ .

Clearly, the origin $(0, 0)$ does not satisfy the inequality $x + 2 y > 3$ .

Hence, the half-plane satisfying $x + 2 y > 3$ does not contain the origin.

Thus, the solution of the constraint $x + 2 y \geq 3$ is the set of all points on the line joining the points $(3, 0)$ and $(1, 1)$ along with the points on the half-plane determined by this line that does not contain the origin.

Step 2

Similarly, the solution of the constraint $2 x + y \geq 3$ consists of all points on the line $2 x + y = 3$ and the half-plane determined by this line satisfying the inequality $2 x + y > 3$ .

Consider the equation $2 x + y = 3$ corresponding to the constraint $2 x + y = 3$ .

When the value $x = 0$ , we have $2 (0) + y = 3 \Rightarrow y = 3$ .

When the value $x = 1$ , we have $2 (1) + y = 3 \Rightarrow y = 1$ .

Clearly, the origin $(0, 0)$ does not satisfy the inequality $2 x + y > 3$ .

Hence, the half-plane satisfying $2 x + y > 3$ does not contain the origin.

Thus, the solution of the constraint $2 x + y \geq 3$ is the set of all points on the line joining the points $(0, 3)$ and $(1, 1)$ along with the points on the half-plane determined by this line that does not contain the origin.

The inequalities $x \geq 0$ and $y \geq 0$ implies that the feasible region lies in the first quadrant.

The intersection of the solutions of these constraints gives the feasible region.

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