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Difficulty: HardSystems of Linear Inequalities in Two Variables

A small factory manufactures two types of toys: wood blocks and toy cars. Let xx represent the number of wood blocks produced daily, and let yy represent the number of toy cars produced daily. The daily production must satisfy the following constraints:

* The total number of toys produced daily cannot exceed 40: x+y≤40x + y \le 40
* Each wood block requires 2 minutes of painting, and each toy car requires 1 minute of painting. The total daily painting time is at most 60 minutes: 2x+y≤602x + y \le 60
* The number of wood blocks produced cannot exceed the number of toy cars produced by more than 15: y≥x−15y \ge x - 15

If the factory must produce a non-negative number of both types of toys, what is the maximum possible number of wood blocks the factory can produce daily?

  1. A
    20
  2. 25Answer
  3. C
    30
  4. D
    40

Answer

The maximum possible number of wood blocks the factory can produce daily is 25.
The correct answer is the value that represents the maximum xx-coordinate within the feasible region. By graphing the inequalities, we find the vertices of the shaded region are (0,0)(0, 0), (15,0)(15, 0), (25,10)(25, 10), (20,20)(20, 20), and (0,40)(0, 40). The maximum xx-value among these vertices is 25, which occurs at the intersection of the painting time constraint and the demand constraint.

Step-by-Step Solution

1
Identify the system of inequalities representing the constraints.
The system of inequalities is:
1) x+y≤40x + y \le 40
2) 2x+y≤602x + y \le 60
3) y≥x−15y \ge x - 15
4) x≥0x \ge 0
5) y≥0y \ge 0
This establishes the boundaries of the feasible region on the coordinate plane.
2
Find the boundary intersection points (vertices of the feasible region) that could maximize xx.
The relevant intersections are:
- The intersection of (1) and (2): x+y=40x + y = 40 and 2x+y=60  ⟹  x=20,y=202x + y = 60 \implies x = 20, y = 20.
- The intersection of (2) and (3): 2x+y=602x + y = 60 and y=x−15  ⟹  2x+(x−15)=60  ⟹  3x=75  ⟹  x=25,y=10y = x - 15 \implies 2x + (x - 15) = 60 \implies 3x = 75 \implies x = 25, y = 10.
- The intersection of (3) and the x-axis (y=0y = 0): 0=x−15  ⟹  x=15,y=00 = x - 15 \implies x = 15, y = 0.
The maximum value of a variable in a bounded linear feasibility region must occur at one of the vertices of the region.
3
Test the vertices to verify they satisfy all inequalities in the system.
- For (20,20)(20, 20): 20+20≤4020 + 20 \le 40 (True), 2(20)+20≤602(20) + 20 \le 60 (True), 20≥20−1520 \ge 20 - 15 (True).
- For (25,10)(25, 10): 25+10=35≤4025 + 10 = 35 \le 40 (True), 2(25)+10=60≤602(25) + 10 = 60 \le 60 (True), 10≥25−1510 \ge 25 - 15 (True).
- For (15,0)(15, 0): 15+0≤4015 + 0 \le 40 (True), 2(15)+0≤602(15) + 0 \le 60 (True), 0≥15−150 \ge 15 - 15 (True).
This ensures that the intersection points actually lie within the feasible region defined by all constraints.
4
Compare the xx-values of the valid vertices to find the maximum.
The candidate xx-values are 20, 25, and 15. The maximum value is 25.
This determines the absolute maximum coordinate value within the bounded region.

Key Concept

Solving systems of linear inequalities by identifying the vertices of the feasible region to optimize a coordinate value.
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Topics

Systems of Linear Inequalities in Two Variables

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