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Difficulty: HardSystems of Linear Inequalities in Two Variables

A company plans to install standard-charging ports and fast-charging ports at its office building. The company must install at least 1515 charging ports in total. Each fast-charging port requires 1212 kilowatts of power, and each standard-charging port requires 44 kilowatts of power. The electrical grid can supply a maximum of 120120 kilowatts of power for these ports. Additionally, the number of standard-charging ports must be at least twice the number of fast-charging ports. What is the maximum number of fast-charging ports the company can install?

Answer: 6

Answer

The maximum number of fast-charging ports the company can install is 6.
Representing the number of fast-charging ports by xx and standard-charging ports by yy yields the system of inequalities: x+y≥15x + y \geq 15, 12x+4y≤12012x + 4y \leq 120, and y≥2xy \geq 2x. The power constraint simplifies to 3x+y≤303x + y \leq 30, or y≤30−3xy \leq 30 - 3x. Combining this with y≥2xy \geq 2x gives 2x≤30−3x2x \leq 30 - 3x, which simplifies to 5x≤305x \leq 30, or x≤6x \leq 6. Substituting x=6x = 6 into the constraints gives y=12y = 12, which satisfies the total port constraint because 6+12=18≥156 + 12 = 18 \geq 15. Therefore, the maximum number of fast-charging ports is 6.

Step-by-Step Solution

1
Define variables and write the system of inequalities representing the constraints.
Let xx be the number of fast-charging ports and yy be the number of standard-charging ports. The constraints are:
1) x+y≥15x + y \geq 15
2) 12x+4y≤12012x + 4y \leq 120
3) y≥2xy \geq 2x
4) x≥0,y≥0x \geq 0, y \geq 0
This translates the word problem constraints into a system of linear inequalities.
2
Simplify the power capacity inequality.
3x+y≤303x + y \leq 30, which can be rewritten as y≤30−3xy \leq 30 - 3x.
Dividing the terms by 4 simplifies the coefficients, making algebraic manipulation easier.
3
Combine the simplified inequality with the charging port ratio constraint to find the upper limit for xx.
Since 2x≤y2x \leq y and y≤30−3xy \leq 30 - 3x, we have 2x≤30−3x2x \leq 30 - 3x.
Adding 3x3x to both sides yields 5x≤305x \leq 30.
Dividing by 5 gives x≤6x \leq 6.
This determines the maximum possible value for the number of fast-charging ports.
4
Verify that the upper limit x=6x = 6 satisfies all system constraints with integer values.
If x=6x = 6, then y≥2(6)=12y \geq 2(6) = 12 and y≤30−3(6)=12y \leq 30 - 3(6) = 12, which means y=12y = 12.
Checking the total port constraint: x+y=6+12=18x + y = 6 + 12 = 18. Since 18≥1518 \geq 15, the point (6,12)(6, 12) satisfies all constraints.
Since the number of ports must be integers, we must confirm that x=6x = 6 yields a valid integer coordinate (6,12)(6, 12) that lies within the feasible region.

Key Concept

Maximizing a variable under a system of linear inequalities in two variables.
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Topics

Systems of Linear Inequalities in Two Variables

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