Math Problem Statement

Each month a store owner Adler can spend at most PHP 1,000,000 on PCs and laptops. A PC costs the store owner PHP 15,000 and a laptop costs him PHP 20,000. Each PC is sold for a profit of PHP 10,000 while a laptop is sold for a profit of PHP 15,000. The store owner estimates that at least 12 PCs but no more than 30 are sold each month. He also estimates that the number of laptops sold is at most half the PCs. How many PCs and laptops should be sold to maximize the profit?

Solution

The problem you've shared is an optimization problem that involves maximizing profit given certain constraints. Here's a step-by-step approach to solve it:

Problem Restatement:

  • Budget Constraint: The store owner can spend at most PHP 1,000,000 on PCs and laptops.
  • Costs:
    • Each PC costs PHP 15,000.
    • Each laptop costs PHP 20,000.
  • Profit:
    • Each PC yields a profit of PHP 10,000.
    • Each laptop yields a profit of PHP 15,000.
  • Sales Constraints:
    • At least 12 PCs must be sold, but no more than 30.
    • The number of laptops sold must be at most half the number of PCs sold.

Variables:

Let xx be the number of PCs sold and yy be the number of laptops sold.

Objective Function (Profit to Maximize):

P(x,y)=10,000x+15,000yP(x, y) = 10,000x + 15,000y

Constraints:

  1. Budget: 15,000x+20,000y1,000,00015,000x + 20,000y \leq 1,000,000
  2. PCs Sold: 12x3012 \leq x \leq 30
  3. Laptops Sold: yx2y \leq \frac{x}{2}

Solution Steps:

  1. Rewrite the Budget Constraint: 15x+20y1000(in thousands of PHP)15x + 20y \leq 1000 \quad \text{(in thousands of PHP)}

  2. Express yy from yx2y \leq \frac{x}{2}: yx2y \leq \frac{x}{2}

  3. Substitute and Solve:

    • Find combinations of xx and yy that satisfy all constraints and maximize the profit function P(x,y)=10x+15yP(x, y) = 10x + 15y (in thousands of PHP).

Solving the Linear Programming Problem:

Let's manually check a few critical points:

  1. x=12x = 12: y122=6y \leq \frac{12}{2} = 6 Budget constraint: 15(12)+20(6)=180+120=300100015(12) + 20(6) = 180 + 120 = 300 \leq 1000 Profit: P(12,6)=10(12)+15(6)=120+90=210 (in thousands of PHP)P(12, 6) = 10(12) + 15(6) = 120 + 90 = 210 \text{ (in thousands of PHP)}

  2. x=30x = 30: y302=15y \leq \frac{30}{2} = 15 Budget constraint: 15(30)+20(15)=450+300=750100015(30) + 20(15) = 450 + 300 = 750 \leq 1000 Profit: P(30,15)=10(30)+15(15)=300+225=525 (in thousands of PHP)P(30, 15) = 10(30) + 15(15) = 300 + 225 = 525 \text{ (in thousands of PHP)}

Since the point (30,15)(30, 15) satisfies all constraints and maximizes the profit, this is the optimal solution.

Final Answer:

  • The store owner should sell 30 PCs and 15 laptops to maximize profit.

Would you like more details on any specific part of this solution or have any other questions?

Related Questions:

  1. How would the solution change if the cost of a laptop increases to PHP 25,000?
  2. What if the store owner could spend PHP 1,200,000 instead of PHP 1,000,000?
  3. How does the minimum requirement of selling 12 PCs affect the overall solution?
  4. Can you generalize the solution for any given budget constraint?
  5. What if the profit per laptop was lower than the profit per PC?

Tip:

When dealing with linear programming problems, always check the constraints carefully, as they can significantly influence the optimal solution.

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Math Problem Analysis

Mathematical Concepts

Linear Programming
Optimization
Inequalities

Formulas

Profit function: P(x, y) = 10,000x + 15,000y
Budget constraint: 15,000x + 20,000y ≤ 1,000,000
Sales constraint: 12 ≤ x ≤ 30
Laptops constraint: y ≤ x / 2

Theorems

Linear Programming Feasibility
Optimization in Bounded Regions

Suitable Grade Level

Grades 10-12