Math Problem Statement

Solve the linear programming problem below using the simplex algorithm. Include all the solution steps. min 𝑍 = 5π‘₯ βˆ’ 4𝑦 subject to π‘₯ + 𝑦 ≀ 5 2π‘₯ βˆ’ 𝑦 = 16 π‘₯ β‰₯ 0 𝑦 ∈ ℝ

Solution

To solve the given linear programming problem using the simplex algorithm, we'll follow the steps systematically. Here is the problem statement:

Objective function:
Minimize Z=5xβˆ’4y\text{Minimize } Z = 5x - 4y

Subject to the constraints:

  1. x+y≀5x + y \leq 5
  2. 2xβˆ’y=162x - y = 16
  3. xβ‰₯0x \geq 0
  4. y∈Ry \in \mathbb{R}

Since yy is not required to be non-negative, we'll have to use a modified simplex approach and carefully handle the equality and inequality constraints.


Step 1: Convert constraints into standard form

1. Convert the inequality into an equation:
For the constraint x+y≀5x + y \leq 5, introduce a slack variable s1β‰₯0s_1 \geq 0: x+y+s1=5x + y + s_1 = 5

2. Write the equality constraint directly:
2xβˆ’y=162x - y = 16

So, the problem now looks like:

Objective function:
Z=5xβˆ’4yZ = 5x - 4y

Subject to:
x + y + s_1 = 5 \tag{1} 2x - y = 16 \tag{2} xβ‰₯0,β€…β€Šs1β‰₯0,β€…β€Šy∈Rx \geq 0, \; s_1 \geq 0, \; y \in \mathbb{R}


Step 2: Set up the initial simplex tableau

We introduce the following variables:

  • xx: Decision variable
  • yy: Decision variable
  • s1s_1: Slack variable
  • ZZ: Objective function value

To convert the minimization problem into a maximization problem, we rewrite ZZ: Maximize Zβ€²=βˆ’Z=βˆ’5x+4y\text{Maximize } Z' = -Z = -5x + 4y

The initial tableau is:

Basicxxyys1s_1RHS
s1s_11115
yy2-1016
Zβ€²Z'-5400

Step 3: Perform simplex iterations

  1. Determine the entering variable:
    The most negative value in the Zβ€²Z' row is βˆ’5-5, corresponding to xx. So, xx will enter the basis.

  2. Determine the leaving variable:
    Perform the ratio test to find the leaving variable:

    • For row 1: 51=5\frac{5}{1} = 5
    • For row 2: 162=8\frac{16}{2} = 8

    Since 5 is the smallest ratio, s1s_1 leaves the basis.

  3. Pivot to form a new tableau:
    Pivot on the element in the xx column of row 1, which is 1.

Basicxxyys1s_1RHS
xx1115
yy0-3-26
Zβ€²Z'09525

Step 4: Check for optimality

Since all the coefficients in the Zβ€²Z' row are now non-negative, the current solution is optimal.

Step 5: Extract the solution

From the final tableau:

  • Basic variables:
    • x=5x = 5
    • y=6y = 6
  • Slack variable:
    • s1=0s_1 = 0

Substitute into the original objective function:

Z=5xβˆ’4y=5(5)βˆ’4(6)=25βˆ’24=1Z = 5x - 4y = 5(5) - 4(6) = 25 - 24 = 1


Final Answer:

The solution to the linear programming problem is:

  • x=5x = 5
  • y=6y = 6
  • Minimum value of ZZ is 1.

Would you like a detailed explanation of any part? Here are some related questions you might find interesting:

  1. How do we determine the entering and leaving variables in the simplex method?
  2. What are slack variables and why are they used in linear programming?
  3. How do we convert a minimization problem into a maximization problem for the simplex method?
  4. What is the significance of the pivot element in the simplex tableau?
  5. Can we solve linear programming problems graphically and how does it compare with the simplex method?

Tip: Always double-check your final tableau for non-negative coefficients in the objective function row to confirm optimality!

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Simplex Algorithm
Optimization

Formulas

Objective function: Z = 5x - 4y
Slack variable: x + y + s_1 = 5
Simplex Tableau Method

Theorems

Simplex Method
Linear Programming Standard Form

Suitable Grade Level

Undergraduate