Math Problem Statement

Using the graphical method, find the values of x and y to solve the linear programming problem: Minimize Z = 3x + 12y, subject to the constraints 5x + y ≤ 32, x + 3y ≥ 12, and x, y ≥ 0.

Solution

To solve the linear programming problem in the uploaded image, let’s break it down step by step based on the problem's requirements:

Problem Overview:

Objective: Minimize Z=3x+12yZ = 3x + 12y
Subject to constraints:

  1. 5x+y325x + y \leq 32
  2. x+3y12x + 3y \geq 12
  3. x,y0x, y \geq 0 (non-negativity constraints)

Steps:

1. Find the x- and y-intercepts for each constraint:

  • For 5x+y=325x + y = 32:

    • Set x=0x = 0: y=32y = 32.
    • Set y=0y = 0: x=6.4x = 6.4 (32/532/5).
  • For x+3y=12x + 3y = 12:

    • Set x=0x = 0: y=4y = 4.
    • Set y=0y = 0: x=12x = 12.

2. Graph the constraints:

  • Plot the lines for 5x+y=325x + y = 32 and x+3y=12x + 3y = 12 using their intercepts.
  • Shade the feasible region that satisfies both constraints:
    • Below 5x+y=325x + y = 32 (due to \leq).
    • Above x+3y=12x + 3y = 12 (due to \geq).
    • Ensure the region is within the first quadrant (non-negativity).

3. Point of Intersection:

  • Solve the equations 5x+y=325x + y = 32 and x+3y=12x + 3y = 12 simultaneously:
    • From 5x+y=325x + y = 32, express yy: y=325xy = 32 - 5x.
    • Substitute y=325xy = 32 - 5x into x+3y=12x + 3y = 12: x+3(325x)=12x + 3(32 - 5x) = 12 x+9615x=12x + 96 - 15x = 12 14x+96=12-14x + 96 = 12 14x=84    x=614x = 84 \implies x = 6
    • Substitute x=6x = 6 into y=325xy = 32 - 5x: y=325(6)=3230=2y = 32 - 5(6) = 32 - 30 = 2
    • Intersection point: (6,2)(6, 2).

4. Corner Points:

  • Corner points of the feasible region:
    1. (0,4)(0, 4) (intercept of x+3y=12x + 3y = 12).
    2. (6,2)(6, 2) (intersection of 5x+y=325x + y = 32 and x+3y=12x + 3y = 12).
    3. (6.4,0)(6.4, 0) (intercept of 5x+y=325x + y = 32).

5. Substitute into the Objective Function:

  • Evaluate Z=3x+12yZ = 3x + 12y for each corner point:
    • At (0,4)(0, 4): Z=3(0)+12(4)=48Z = 3(0) + 12(4) = 48.
    • At (6,2)(6, 2): Z=3(6)+12(2)=18+24=42Z = 3(6) + 12(2) = 18 + 24 = 42.
    • At (6.4,0)(6.4, 0): Z=3(6.4)+12(0)=19.2Z = 3(6.4) + 12(0) = 19.2.

6. Formulate the Decision:

  • Minimum ZZ occurs at (6.4,0)(6.4, 0) with Z=19.2Z = 19.2.
  • Decision:
    • x=6.4x = 6.4,
    • y=0y = 0,
    • Minimum Z=19.2\text{Minimum } Z = 19.2.

Would you like a detailed graph or further explanation on any step?

Related Questions:

  1. How can you graph the feasible region manually?
  2. Why does the intersection point help solve the problem?
  3. How would the solution change if the objective was to maximize ZZ?
  4. Can we use another method like the Simplex Method to solve this?
  5. What are practical scenarios where linear programming is used?

Tip:

Always confirm that your solution is in the feasible region by substituting back into all constraints.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Graphical Method
Optimization

Formulas

Objective Function: Z = 3x + 12y
Constraints: 5x + y ≤ 32, x + 3y ≥ 12, x, y ≥ 0

Theorems

Feasible Region
Linear Programming Optimization

Suitable Grade Level

College/Advanced High School