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Difficulty: MediumLinear and Exponential Growth

A local utility company is evaluating two different methods, Method A and Method B, to reduce electricity distribution losses. The table below shows the modeled annual electricity losses, in megawatt-hours (MWh), for each method over the first two years (t=0t = 0 to t=2t = 2), where tt is the number of years since the evaluation began.

Year (tt)Method A losses (MWh)Method B losses (MWh)
08,0008,000
17,4007,200
26,8006,480

One method is modeled by a linear function, and the other method is modeled by an exponential function. What is the difference, in MWh, between the modeled annual electricity losses for the two methods at t=3t = 3 years?

  1. 368Answer
  2. B
    600
  3. C
    732
  4. D
    1,800

Answer

368
To find the difference between the modeled annual electricity losses at t=3t = 3, we first identify the type of model for each method. For Method A, the losses decrease by a constant amount of 600 MWh600\text{ MWh} each year (8,000−7,400=6008,000 - 7,400 = 600 and 7,400−6,800=6007,400 - 6,800 = 600), which indicates a linear relationship. Thus, the model for Method A is LA(t)=8,000−600tL_A(t) = 8,000 - 600t. At t=3t = 3, the losses for Method A are LA(3)=8,000−600(3)=6,200 MWhL_A(3) = 8,000 - 600(3) = 6,200\text{ MWh}. For Method B, the losses decrease by a constant percentage of 10%10\% each year, meaning each year's value is 90%90\% of the previous year's value (7,2008,000=0.9\frac{7,200}{8,000} = 0.9 and (6,4807,200=0.9\frac{6,480}{7,200} = 0.9). This indicates an exponential relationship with a decay factor of 0.90.9. Thus, the model for Method B is LB(t)=8,000(0.9)tL_B(t) = 8,000(0.9)^t. At t=3t = 3, the losses for Method B are LB(3)=8,000(0.9)3=8,000(0.729)=5,832 MWhL_B(3) = 8,000(0.9)^3 = 8,000(0.729) = 5,832\text{ MWh}. The difference between the modeled losses is 6,200−5,832=368 MWh6,200 - 5,832 = 368\text{ MWh}.

Step-by-Step Solution

1
Identify the type of model for Method A by examining its rate of change.
Method A decreases by a constant value of 600 MWh600\text{ MWh} each year (8,000−7,400=6008,000 - 7,400 = 600 and 7,400−6,800=6007,400 - 6,800 = 600), which means it is a linear function.
Linear functions have a constant first difference (constant rate of change).
2
Identify the type of model for Method B by examining its ratios.
Method B decreases by a constant ratio of 0.90.9 each year (7,2008,000=0.9\frac{7,200}{8,000} = 0.9 and (6,4807,200=0.9\frac{6,480}{7,200} = 0.9), which means it is an exponential function.
Exponential functions have a constant ratio between consecutive terms.
3
Formulate the equations for both models.
LA(t)=8,000−600tL_A(t) = 8,000 - 600t and LB(t)=8,000(0.9)tL_B(t) = 8,000(0.9)^t.
These equations represent the linear decrease and exponential decay starting from 8,0008,000.
4
Calculate the modeled losses for both methods at t=3t = 3.
LA(3)=8,000−600(3)=6,200 MWhL_A(3) = 8,000 - 600(3) = 6,200\text{ MWh} and LB(3)=8,000(0.9)3=8,000(0.729)=5,832 MWhL_B(3) = 8,000(0.9)^3 = 8,000(0.729) = 5,832\text{ MWh}.
To find the values at t=3t = 3, substitute 33 for tt in both equations.
5
Calculate the difference between the two values.
6,200−5,832=368 MWh6,200 - 5,832 = 368\text{ MWh}.
Subtract the smaller loss value from the larger loss value to find the difference.

Key Concept

Identifying and modeling linear growth/decay (constant change per unit time) versus exponential growth/decay (constant percent change per unit time) from data tables.
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Topics

Linear and Exponential Growth

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