Math Problem Statement

πŸ“Œ Question: A company produces two types of products, P and Q. The company has 24 hours of labor available. P requires 4 hours per unit, while Q requires 3 hours per unit. The company can manufacture at most 6 units of P and at most 8 units of Q. The profit per unit is β‚Ή60 for P and β‚Ή50 for Q.

Formulate and solve graphically to maximize profit.

Solution

Step 1: Define Variables

Let:

  • xx be the number of units of product PP.
  • yy be the number of units of product QQ.

Step 2: Objective Function

The company wants to maximize profit, which is given by:

Z=60x+50yZ = 60x + 50y

Step 3: Constraints

  1. Labor Constraint: Each unit of PP requires 4 hours, and each unit of QQ requires 3 hours. The company has 24 hours available:

    4x+3y≀244x + 3y \leq 24

  2. Production Limits:

    • At most 6 units of PP:

      x≀6x \leq 6

    • At most 8 units of QQ:

      y≀8y \leq 8

  3. Non-Negativity Constraints: Since the number of units cannot be negative,

    xβ‰₯0,yβ‰₯0x \geq 0, \quad y \geq 0

Step 4: Graphical Solution

We graph the feasible region by plotting the constraints.

Plotting the Constraint 4x+3y≀244x + 3y \leq 24

  • Find intercepts:

    • Set x=0x = 0: 3y=24β‡’y=83y = 24 \Rightarrow y = 8 β†’ Point (0,8)(0,8)
    • Set y=0y = 0: 4x=24β‡’x=64x = 24 \Rightarrow x = 6 β†’ Point (6,0)(6,0)
  • Other Constraints:

    • The vertical line x=6x = 6
    • The horizontal line y=8y = 8
    • The region is bounded in the first quadrant.

Finding Corner Points

The feasible region is bounded by:

  1. (0,0) – No production.

  2. (6,0) – Producing only P.

  3. (0,8) – Producing only Q.

  4. Intersection of 4x+3y=244x + 3y = 24 with x=6x = 6

    • Substituting x=6x = 6 in 4(6)+3y=244(6) + 3y = 24:

      24+3y=24β‡’y=024 + 3y = 24 \Rightarrow y = 0

      So, the point is (6,0)(6,0), which was already considered.

  5. Intersection of 4x+3y=244x + 3y = 24 with y=8y = 8

    • Substituting y=8y = 8 in 4x+3(8)=244x + 3(8) = 24:

      4x+24=24β‡’4x=0β‡’x=04x + 24 = 24 \Rightarrow 4x = 0 \Rightarrow x = 0

      So, the point is (0,8)(0,8), which was already considered.

Thus, the corner points are: (0,0)(0,0), (6,0)(6,0), and (0,8)(0,8).

Step 5: Calculate Profit at Each Corner

  • At (0,0): Z=60(0)+50(0)=0Z = 60(0) + 50(0) = 0
  • At (6,0): Z=60(6)+50(0)=360Z = 60(6) + 50(0) = 360
  • At (0,8): Z=60(0)+50(8)=400Z = 60(0) + 50(8) = 400

Step 6: Conclusion

The maximum profit is β‚Ή400, achieved by producing 0 units of P and 8 units of Q.


Would you like a graphical representation?

Here are some related questions:

  1. What happens if the labor hours increase from 24 to 30?
  2. How would the solution change if profit per unit of Q increased to β‚Ή60?
  3. What if an additional constraint, like a budget limit, is introduced?
  4. How do we solve this problem using the Simplex Method instead of graphically?
  5. How would we interpret the shadow price of labor in this case?

πŸ’‘ Tip: In graphical solutions, always check the feasible region before evaluating corner points.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Graphical Method
Optimization

Formulas

Objective Function: Z = 60x + 50y
Labor Constraint: 4x + 3y ≀ 24
Production Constraints: x ≀ 6, y ≀ 8
Non-Negativity Constraints: x β‰₯ 0, y β‰₯ 0

Theorems

Fundamental Theorem of Linear Programming
Corner Point Theorem

Suitable Grade Level

Grades 10-12