Math Problem Statement

Solve the LP problem. If no optimal solution exists, indicate whether the feasible region is empty or the objective function is unbounded. (Enter EMPTY if the region is empty. Enter UNBOUNDED if the function is unbounded.)

Minimize c = x + 2y subject to

x + 9y

26

3x + y

26

x ≥ 0, y ≥ 0.

c= (xy)

Solution

We are asked to solve the following linear programming (LP) problem:

Objective: Minimize c=x+2yc = x + 2y

Subject to the constraints:

  1. x+9y26x + 9y \geq 26
  2. 3x+y263x + y \geq 26
  3. x0x \geq 0
  4. y0y \geq 0

Step 1: Analyze the constraints graphically

Each inequality represents a half-plane in the xyxy-plane. We need to find the region where all the constraints hold simultaneously.

1. x+9y26x + 9y \geq 26

This can be rewritten as: y26x9y \geq \frac{26 - x}{9} This is a line with slope 19-\frac{1}{9} and y-intercept 2692.89\frac{26}{9} \approx 2.89.

2. 3x+y263x + y \geq 26

This can be rewritten as: y263xy \geq 26 - 3x This is a line with slope 3-3 and y-intercept 26.

3. x0x \geq 0

This represents the region to the right of the yy-axis.

4. y0y \geq 0

This represents the region above the xx-axis.

Step 2: Find the intersection of constraints

The feasible region is where all the constraints overlap.

Intersection of x+9y=26x + 9y = 26 and 3x+y=263x + y = 26:

We solve these two equations to find the point of intersection.

  1. From the first equation x+9y=26x + 9y = 26, solve for xx: x=269yx = 26 - 9y

  2. Substitute this into the second equation 3x+y=263x + y = 26: 3(269y)+y=263(26 - 9y) + y = 26 7827y+y=2678 - 27y + y = 26 26y=52-26y = -52 y=2y = 2

  3. Substitute y=2y = 2 back into x=269yx = 26 - 9y: x=269(2)=2618=8x = 26 - 9(2) = 26 - 18 = 8

So, the point of intersection is (8,2)(8, 2).

Step 3: Check the boundaries

We now check the boundaries of the feasible region by considering the constraints at extreme points.

  • At y=0y = 0:
    • From x+9y=26x + 9y = 26, we get x=26x = 26, giving the point (26,0)(26, 0).
    • From 3x+y=263x + y = 26, we get x=2638.67x = \frac{26}{3} \approx 8.67, giving the point (8.67,0)(8.67, 0).

The feasible region lies between these extreme points, including the point of intersection (8,2)(8, 2).

Step 4: Evaluate the objective function at corner points

Now, evaluate c=x+2yc = x + 2y at the corner points of the feasible region.

  1. At (26,0)(26, 0): c=26+2(0)=26c = 26 + 2(0) = 26

  2. At (8,2)(8, 2): c=8+2(2)=8+4=12c = 8 + 2(2) = 8 + 4 = 12

  3. At (8.67,0)(8.67, 0): c=8.67+2(0)=8.67c = 8.67 + 2(0) = 8.67

Step 5: Conclusion

The minimum value of cc occurs at the point (8,2)(8, 2) with c=12c = 12.

Thus, the optimal solution is: 12\boxed{12}

Would you like more details or further explanations?


Here are 5 relative questions to deepen your understanding:

  1. How do we determine whether the objective function is unbounded in an LP problem?
  2. What is the geometric significance of the feasible region in linear programming?
  3. How does changing the objective function affect the location of the optimal solution?
  4. Can an LP problem have multiple optimal solutions? Under what conditions?
  5. What methods can be used to solve LP problems besides the graphical approach?

Tip: In LP, the feasible region is always a convex set, which means the optimal solution is found at a vertex (corner point) if it exists.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Graphical Method
Inequalities
Optimization

Formulas

Objective function: c = x + 2y
Constraint 1: x + 9y ≥ 26
Constraint 2: 3x + y ≥ 26
Non-negativity constraints: x ≥ 0, y ≥ 0

Theorems

Fundamental Theorem of Linear Programming

Suitable Grade Level

College Level or Advanced High School