Math Problem Statement

Use the technique developed in this section to solve the minimization problem. Minimize C = 10x + y subject to 4x + y ≥ 15 x + 2y ≥ 11 x ≥ 2 x ≥ 0, y ≥ 0   The minimum is C =

Incorrect: Your answer is incorrect. at (x, y) =

Incorrect: Your answer is incorrect.

Solution

To solve this linear programming problem, we can use the method of finding the feasible region, determining the corner points, and evaluating C=10x+yC = 10x + y at those points to find the minimum value.

Step 1: Rewrite the Inequalities

We have:

  1. 4x+y154x + y \geq 15
  2. x+2y11x + 2y \geq 11
  3. x2x \geq 2
  4. x0x \geq 0 and y0y \geq 0

To graphically solve this, we will treat these inequalities as equalities to find where the lines intersect, which will give us the vertices (corner points) of the feasible region.

Step 2: Find the Intersection Points

1. Intersection of 4x+y=154x + y = 15 and x+2y=11x + 2y = 11:

  • Solve these equations simultaneously: 4x+y=154x + y = 15 x+2y=11x + 2y = 11
  • Multiply the second equation by 4 to eliminate xx: 4x+y=154x + y = 15 4x+8y=444x + 8y = 44
  • Subtract the first equation from the second: 7y=29y=2974.147y = 29 \Rightarrow y = \frac{29}{7} \approx 4.14
  • Substitute y=297y = \frac{29}{7} into x+2y=11x + 2y = 11: x+2297=11x + 2 \cdot \frac{29}{7} = 11 x+587=11x + \frac{58}{7} = 11 x=11587=1972.71x = 11 - \frac{58}{7} = \frac{19}{7} \approx 2.71
  • So, the intersection point is approximately (197,297)\left( \frac{19}{7}, \frac{29}{7} \right).

2. Intersection of 4x+y=154x + y = 15 and x=2x = 2:

  • Substitute x=2x = 2 into 4x+y=154x + y = 15: 4(2)+y=154(2) + y = 15 8+y=15y=78 + y = 15 \Rightarrow y = 7
  • So, the intersection point is (2,7)(2, 7).

3. Intersection of x+2y=11x + 2y = 11 and x=2x = 2:

  • Substitute x=2x = 2 into x+2y=11x + 2y = 11: 2+2y=112 + 2y = 11 2y=9y=92=4.52y = 9 \Rightarrow y = \frac{9}{2} = 4.5
  • So, the intersection point is (2,4.5)(2, 4.5).

Step 3: Evaluate C=10x+yC = 10x + y at Each Vertex

Now we evaluate C=10x+yC = 10x + y at each vertex to find the minimum.

  1. At (197,297)\left( \frac{19}{7}, \frac{29}{7} \right): C=10197+297=1907+297=219731.29C = 10 \cdot \frac{19}{7} + \frac{29}{7} = \frac{190}{7} + \frac{29}{7} = \frac{219}{7} \approx 31.29

  2. At (2,7)(2, 7): C=102+7=20+7=27C = 10 \cdot 2 + 7 = 20 + 7 = 27

  3. At (2,4.5)(2, 4.5): C=102+4.5=20+4.5=24.5C = 10 \cdot 2 + 4.5 = 20 + 4.5 = 24.5

Step 4: Determine the Minimum Value

The minimum value of C=10x+yC = 10x + y is 24.524.5, which occurs at (x,y)=(2,4.5)(x, y) = (2, 4.5).

Answer

The minimum value is: C=24.5 at (x,y)=(2,4.5)C = 24.5 \text{ at } (x, y) = (2, 4.5)

Would you like more details or have any further questions?


Related Questions:

  1. How can we use graphical methods to solve other linear programming problems?
  2. How can we confirm the feasible region visually using graphing tools?
  3. What is the significance of the corner-point principle in linear programming?
  4. Are there other methods besides the graphical approach to solve linear programming problems?
  5. How does changing the coefficients in C=10x+yC = 10x + y affect the solution?

Tip: In linear programming, always check each vertex of the feasible region for minimum and maximum values, as the optimal solution lies at one of the vertices.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Inequalities
Optimization

Formulas

C = 10x + y
4x + y = 15
x + 2y = 11

Theorems

Corner-Point Principle

Suitable Grade Level

Grades 11-12