Math Problem Statement

Consider the following linear program on two variables x and y: Maximize 3x + 5y Subject to x+2y <= 5; 2x + 2y <= 7; 4x + 3y <= 12; x, y >= 0. We add the condition to this linear program that the variables x and y cannot take values between 1 and 2, i.e., each of them is allowed to take values of 1 or 2, but not 1.23. Including the added condition, what would be the optimal objective function value and what would be the values of x and y in an optimal solution? Answer: Objective function value = ?, x = ?, y= ?

Solution

To solve this modified linear programming problem, we incorporate the condition that xx and yy can only take integer values (since values between 1 and 2 are disallowed, and fractional values of xx and yy are not allowed). This transforms the problem into an integer linear programming (ILP) problem.

The problem can be written as:

Objective function: Maximize 3x+5y3x + 5y

Constraints:

  1. x+2y5x + 2y \leq 5
  2. 2x+2y72x + 2y \leq 7
  3. 4x+3y124x + 3y \leq 12
  4. x,y0x, y \geq 0
  5. xx and yy must be integers.

We will now calculate the optimal values of xx and yy using integer programming methods.

Step 1: Determine feasible integer solutions

We systematically evaluate feasible integer values of xx and yy by satisfying the constraints:

  • From x+2y5x + 2y \leq 5, x52yx \leq 5 - 2y.
  • From 2x+2y72x + 2y \leq 7, x+y3.5x + y \leq 3.5, or x+y3x + y \leq 3 for integer solutions.
  • From 4x+3y124x + 3y \leq 12, x30.75yx \leq 3 - 0.75y.

Feasible pairs for xx and yy:

We test values systematically:

  1. y=0y = 0:
    • x5x \leq 5, x3.5x \leq 3.5, x3x \leq 3x=0,1,2,3x = 0, 1, 2, 3.
  2. y=1y = 1:
    • x3x \leq 3, x2.5x \leq 2.5, x2.25x \leq 2.25x=0,1,2x = 0, 1, 2.
  3. y=2y = 2:
    • x1x \leq 1, x1.5x \leq 1.5, x1.75x \leq 1.75x=0,1x = 0, 1.
  4. y=3y = 3: Not feasible, as it violates constraints.

Step 2: Evaluate the objective function

We evaluate 3x+5y3x + 5y for all feasible integer pairs:

  • (x,y)=(0,0)(x, y) = (0, 0): 3(0)+5(0)=03(0) + 5(0) = 0
  • (x,y)=(1,0)(x, y) = (1, 0): 3(1)+5(0)=33(1) + 5(0) = 3
  • (x,y)=(2,0)(x, y) = (2, 0): 3(2)+5(0)=63(2) + 5(0) = 6
  • (x,y)=(3,0)(x, y) = (3, 0): 3(3)+5(0)=93(3) + 5(0) = 9
  • (x,y)=(0,1)(x, y) = (0, 1): 3(0)+5(1)=53(0) + 5(1) = 5
  • (x,y)=(1,1)(x, y) = (1, 1): 3(1)+5(1)=83(1) + 5(1) = 8
  • (x,y)=(2,1)(x, y) = (2, 1): 3(2)+5(1)=113(2) + 5(1) = 11
  • (x,y)=(0,2)(x, y) = (0, 2): 3(0)+5(2)=103(0) + 5(2) = 10
  • (x,y)=(1,2)(x, y) = (1, 2): 3(1)+5(2)=133(1) + 5(2) = 13.

Step 3: Identify the optimal solution

The maximum value of the objective function is 13, achieved at (x,y)=(1,2)(x, y) = (1, 2).

Final Answer:

  • Objective function value: 13
  • xx: 1
  • yy: 2

Would you like further clarification or a step-by-step breakdown of any part of this process?


Expanding Questions:

  1. What are the differences between linear programming and integer programming?
  2. How would the solution change if xx and yy were allowed to take fractional values?
  3. What are practical examples of problems that use integer programming?
  4. Can you explain how the feasible region is constructed graphically for this problem?
  5. What optimization methods can be used for more complex integer programming problems?

Tip: Always ensure the feasibility of solutions by checking all constraints when solving optimization problems.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Integer Programming
Optimization

Formulas

Objective function: Maximize 3x + 5y
Constraint inequalities: x + 2y <= 5, 2x + 2y <= 7, 4x + 3y <= 12, x, y >= 0

Theorems

Fundamental theorem of linear programming

Suitable Grade Level

Undergraduate level or advanced high school (Grades 11-12)