The problem describes a new electricity billing system. There is a fixed installation fee of K15. For the first 200 units (kilowatt hours), the charge is 50 toea per unit. For units exceeding 200, the charge is 60 toea per unit. The task is to construct a formula for the total charging function, which includes VAT.

AlgebraPiecewise FunctionsLinear EquationsModelingWord Problem
2025/6/4

1. Problem Description

The problem describes a new electricity billing system. There is a fixed installation fee of K
1

5. For the first 200 units (kilowatt hours), the charge is 50 toea per unit. For units exceeding 200, the charge is 60 toea per unit. The task is to construct a formula for the total charging function, which includes VAT.

2. Solution Steps

Let xx be the number of units (kilowatt hours) consumed.
Let C(x)C(x) be the total charge in toea.
Case 1: 0x2000 \le x \le 200
The cost is the installation fee plus the unit cost.
C(x)=1500+50xC(x) = 1500 + 50x
Case 2: x>200x > 200
The cost is the installation fee plus the cost of the first 200 units plus the cost of the remaining units. The cost of the first 200 units is 50×200=1000050 \times 200 = 10000 toea. The number of units exceeding 200 is x200x - 200. The cost of these exceeding units is 60(x200)60(x - 200).
C(x)=1500+10000+60(x200)C(x) = 1500 + 10000 + 60(x - 200)
C(x)=11500+60x12000C(x) = 11500 + 60x - 12000
C(x)=60x500C(x) = 60x - 500
Therefore, the charging function is:
$C(x) = \begin{cases}
1500 + 50x, & 0 \le x \le 200 \\
60x - 500, & x > 200
\end{cases}$

3. Final Answer

The charging function is:
$C(x) = \begin{cases}
1500 + 50x, & 0 \le x \le 200 \\
60x - 500, & x > 200
\end{cases}$
where C(x)C(x) is the total charge in toea and xx is the number of units consumed.