Math Problem Statement

Maximize Z = 4x + 6y subject to the constraints 2x + y <= 8, x + 2y <= 10, and x, y >= 0.

Solution

The problem you've provided is a linear programming problem. Here's the formulation:

Objective:

Maximize Z=4x+6yZ = 4x + 6y

Subject to the constraints:

  1. 2x+y82x + y \leq 8
  2. x+2y10x + 2y \leq 10
  3. x0x \geq 0
  4. y0y \geq 0

This problem aims to find the values of xx and yy that maximize Z=4x+6yZ = 4x + 6y while satisfying the given constraints.

Solution process:

  1. Graph the constraints: Plot the inequalities 2x+y82x + y \leq 8, x+2y10x + 2y \leq 10, along with the non-negativity conditions x0x \geq 0 and y0y \geq 0. These inequalities form a feasible region.

  2. Find the corner points: The maximum value of the objective function will occur at one of the corner points of the feasible region. We'll determine these points by solving the system of inequalities.

  3. Evaluate the objective function at each corner point to find the maximum value of Z=4x+6yZ = 4x + 6y.

Let's go through this systematically:

Step 1: Find the corner points

Solve the system of equations formed by the intersection of the boundary lines of the constraints:

  1. 2x+y=82x + y = 8
  2. x+2y=10x + 2y = 10
  • Intersection of 2x+y=82x + y = 8 and x+2y=10x + 2y = 10:
    • Solve the first equation for yy: y=82xy = 8 - 2x.
    • Substitute into the second equation: x+2(82x)=10x+164x=103x=6x=2.x + 2(8 - 2x) = 10 \Rightarrow x + 16 - 4x = 10 \Rightarrow -3x = -6 \Rightarrow x = 2.
    • Substitute x=2x = 2 into y=82xy = 8 - 2x: y=84=4.y = 8 - 4 = 4. So, the intersection point is (2,4)(2, 4).

Other corner points come from the intersections with the axes:

  • Intersection with xx-axis: Set y=0y = 0 in 2x+y=82x + y = 8, yielding x=4x = 4. So, one corner point is (4,0)(4, 0).

  • Intersection with yy-axis: Set x=0x = 0 in x+2y=10x + 2y = 10, yielding y=5y = 5. So, another corner point is (0,5)(0, 5).

Step 2: Evaluate the objective function at each corner point

  • At (4,0)(4, 0): Z=4(4)+6(0)=16Z = 4(4) + 6(0) = 16.
  • At (0,5)(0, 5): Z=4(0)+6(5)=30Z = 4(0) + 6(5) = 30.
  • At (2,4)(2, 4): Z=4(2)+6(4)=8+24=32Z = 4(2) + 6(4) = 8 + 24 = 32.

Step 3: Conclusion

The maximum value of Z=32Z = 32 occurs at the point (2,4)(2, 4).

Would you like further details on any part of the solution?

Here are some related questions to expand understanding:

  1. What are the graphical interpretations of linear programming constraints?
  2. How do you know which points in the feasible region will yield the maximum or minimum value?
  3. Can a linear programming problem have more than one optimal solution?
  4. What happens if the feasible region is unbounded?
  5. How would the solution change if the objective function was to minimize ZZ?

Tip: In linear programming, the optimal solution always lies at a vertex (corner point) of the feasible region.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Graphical Method

Formulas

Objective Function: Z = 4x + 6y
Constraints: 2x + y <= 8, x + 2y <= 10, x >= 0, y >= 0

Theorems

Fundamental Theorem of Linear Programming

Suitable Grade Level

Grade 11-12 or College-level