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Difficulty: MediumLinear Functions and Graphs

An environmental scientist is monitoring the water level of a reservoir during a dry season. The water level, L(d)L(d), in meters, can be modeled by a linear function of the number of days, dd, since the start of the dry season. On day 12, the water level was 30 meters, and on day 20, the water level was 26 meters. If the water level continues to decrease at this constant rate, on which day will the water level be exactly 18 meters?

  1. A
    Day 18
  2. B
    Day 24
  3. Day 36Answer
  4. D
    Day 40

Answer

Day 36
To find the day when the water level is exactly 18 meters, we first find the constant rate of change (slope) using the given points (12,30)(12, 30) and (20,26)(20, 26). The slope is m=26−3020−12=−0.5m = \frac{26 - 30}{20 - 12} = -0.5 meters per day. Using the point-slope form with (12,30)(12, 30), we get L(d)−30=−0.5(d−12)L(d) - 30 = -0.5(d - 12), which simplifies to L(d)=−0.5d+36L(d) = -0.5d + 36. Substituting 18 for L(d)L(d) gives 18=−0.5d+3618 = -0.5d + 36. Subtracting 36 from both sides results in −18=−0.5d-18 = -0.5d, and dividing by −0.5-0.5 yields d=36d = 36. Thus, the correct answer is Day 36.

Step-by-Step Solution

1
Calculate the constant rate of change (slope, mm) using the points (12,30)(12, 30) and (20,26)(20, 26).
m=26−3020−12=−48=−0.5m = \frac{26 - 30}{20 - 12} = \frac{-4}{8} = -0.5 meters per day.
A linear function has a constant rate of change, which is determined by the slope formula: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1}.
2
Write the equation of the linear model in point-slope form and solve for the standard form.
L(d)−30=−0.5(d−12)⇒L(d)=−0.5d+36L(d) - 30 = -0.5(d - 12) \Rightarrow L(d) = -0.5d + 36.
Using the point-slope formula y−y1=m(x−x1)y - y_1 = m(x - x_1) with the point (12,30)(12, 30) and slope −0.5-0.5 defines the function for any day dd.
3
Substitute L(d)=18L(d) = 18 into the equation and solve for dd.
18=−0.5d+36⇒−18=−0.5d⇒d=3618 = -0.5d + 36 \Rightarrow -18 = -0.5d \Rightarrow d = 36.
Setting the dependent variable to the target water level of 18 meters yields the corresponding day.

Key Concept

Determining and evaluating linear equations from two given coordinate points

Alternative Method

Instead of constructing the entire equation, you can analyze the constant rate of change directly. The water level decreases by 44 meters (from 30 to 26) over 88 days (from day 12 to day 20), which means the rate of decrease is 0.50.5 meters per day. To drop from 26 meters (on day 20) to 18 meters requires a further decrease of 26−18=826 - 18 = 8 meters. At a rate of 0.50.5 meters per day, dropping 8 meters will take 80.5=16\frac{8}{0.5} = 16 days. Adding these 16 days to day 20 yields day 36.
Estimated Time:1m 30s
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Topics

Linear Functions and Graphs

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