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Lesson plan Grade 12 Math

Fetterly's Rate Increase: When Customers Use Less Water

A 55-minute grade 12 lesson where students model a water utility's revenue against a demand response and solve a quadratic for the rate increase that hits a target.

Objectives

  1. Students will write utility revenue as a product of a rising price and a falling quantity.
  2. Students will solve a quadratic equation for the rate increase that meets a revenue target.
  3. Students will judge which of two algebraic roots is viable in the modeling context.

Vocabulary

Revenue requirement
The total amount of money a utility must collect from customers each year to run the water system safely. Rates are set so that all the bills together add up to this amount.

Also: demand response, elasticity, viable solution, rate per thousand gallons

Teacher overview

A water utility that needs more revenue raises its rate, and its customers respond by using less water, so the revenue that arrives is smaller than the arithmetic of the rate increase alone predicts. Utilities describe that response with a demand elasticity, and setting a rate to hit a revenue target becomes a quadratic problem rather than a proportion. This lesson uses the fictional Fetterly Water District, whose board wants 12 percent more revenue and whose analyst estimates that every 1 percent increase in the rate lowers average use by 0.30 percent. The misconception the lesson exists to correct is the belief that a 12 percent rate increase produces 12 percent more revenue, an assumption that leaves the fictional district about $16,700 short in the first month.

Procedure

Launch (7 min): Give the class the fictional Fetterly figures, which are 12,400 accounts using an average of 6,200 gallons a month at $5.40 per thousand gallons, and ask what rate increase would raise revenue by 12 percent. Take the immediate answer, which is almost always 12 percent, and write it on the board without correcting it. Then ask what customers do when a bill goes up.

Direct instruction (14 min): Introduce the demand response as a stated elasticity of negative 0.30 and define it precisely: a 1 percent rate increase lowers average use by 0.30 percent. Build the revenue expression with the class as current revenue times (1 + r) times (1 - 0.30r), where r is the fractional rate increase, and name each factor. Expand the product on the board so students see the quadratic term appear, and explain that the term exists because both price and quantity move at once.

Guided practice (11 min): Work the naive case together. At r = 0.12 the new rate is $6.048 and average use falls to 5,976.8 gallons, so revenue reaches about $448,231 against a target of about $464,970. Ask pairs to state, in one sentence, why the revenue gain is smaller than the 12 percent rate increase less the 3.6 percent drop in use. Take two or three answers before moving on.

Independent practice (18 min): Students complete questions 1 through 10 on the handout using the same fictional Fetterly figures. Watch for two errors: writing the demand factor as (1 - 0.30) instead of (1 - 0.30r), and dropping the negative sign when arranging the quadratic in standard form.

Closure and formative check (5 min): On an exit slip, students state which root of the quadratic they used and give one sentence explaining why the other root was rejected. Collect the slips and count how many describe the rejected root in terms of the model's range rather than simply calling it too large. If fewer than half do, open the next session with a graph of revenue against r showing where the model stops meaning anything.

Fetterly's Rate Increase: When Customers Use Less Water — Grade 12 Math
Name: ____________________________ Date: ______________

Student Handout

The fictional Fetterly Water District serves 12,400 accounts. Each account uses an average of 6,200 gallons per month, and the district charges $5.40 per thousand gallons. The district's analyst estimates that each 1 percent increase in the rate lowers average monthly use by 0.30 percent. Let r stand for the fractional rate increase, so r = 0.12 means a 12 percent increase.

  1. Compute the district's current monthly revenue.
  2. The board wants monthly revenue 12 percent higher than it is now. What is the target monthly revenue?
  3. If the rate rises 12 percent, what is the new average monthly use per account, in gallons? Round to the nearest tenth of a gallon.
  4. If the rate rises 12 percent, what is the new rate per thousand gallons?
  5. Using your answers to questions 3 and 4, compute the monthly revenue the district would actually collect after a 12 percent rate increase. Round to the nearest dollar.
  6. By how many dollars does the result in question 5 fall short of the target in question 2? Round to the nearest dollar.
  7. Revenue after an increase of r equals the current revenue times (1 + r) times (1 - 0.30r). Setting that equal to 1.12 times current revenue and simplifying gives a quadratic in r. Write it in the standard form ar^2 + br + c = 0 with a positive.
  8. Solve your quadratic and report the smaller root, rounded to three decimal places.
  9. Using the smaller root, what rate per thousand gallons should the district set? Round to the nearest cent.
  10. The quadratic has a second root of about 2.147. Explain in two or three sentences why the district cannot use it.

From Water for Schools (waterforschools.org). This material is in the public domain (CC0) and may be copied, edited, and shared freely.

Answer Key

  1. Answer: $415,152.
    Each account is billed 6.2 thousand gallons at $5.40, or $33.48 per month, and 12,400 accounts times $33.48 gives $415,152.

  2. Answer: $464,970.
    Twelve percent above $415,152 is 415,152 times 1.12, or about $464,970.24, which rounds to $464,970.

  3. Answer: 5,976.8 gallons.
    A 12 percent rate increase lowers use by 0.30 times 12 percent, or 3.6 percent, so use becomes 6,200 times 0.964, which is 5,976.8 gallons.

  4. Answer: $6.048 per thousand gallons.
    The current rate of $5.40 times 1.12 is $6.048 per thousand gallons.

  5. Answer: $448,231.
    Each account is billed 5.9768 thousand gallons at $6.048, or about $36.148, and 12,400 accounts gives about $448,231.

  6. Answer: $16,739.
    The target of $464,970.24 less the collected $448,231.31 leaves a shortfall of about $16,738.93, which rounds to $16,739.

  7. Answer: 0.3r^2 - 0.7r + 0.12 = 0.
    Expanding (1 + r)(1 - 0.30r) gives 1 + 0.7r - 0.3r^2. Setting that equal to 1.12 and moving every term to the left with a positive leading coefficient gives 0.3r^2 - 0.7r + 0.12 = 0.

  8. Answer: 0.186.
    The quadratic formula gives r = (0.7 minus the square root of 0.346) divided by 0.6. The square root of 0.346 is about 0.58822, so r is about 0.18630, which rounds to 0.186.

  9. Answer: $6.41 per thousand gallons.
    The rate becomes 5.40 times 1.18630, or about $6.4060, which rounds to $6.41 per thousand gallons. Average use falls to about 5,853 gallons, and the revenue check returns about $464,970.

  10. Answer: A 214.7 percent rate increase falls far outside the range where the estimated demand response holds.
    The elasticity was estimated from small changes around the current rate, and applying it to a rate more than triple the present one predicts a 64 percent drop in use with no basis in the data behind the estimate. The larger root solves the equation but does not describe anything the district could do, so it is not a viable solution.

Differentiation

Support

Give these students the revenue expression already written out for the fictional Fetterly district, with the price factor and the quantity factor labeled, so their work starts at the expansion rather than at the modeling. Provide a filled-in table of revenue at r values of 0.05, 0.10, 0.15, and 0.20, and let them locate the target between two rows before solving the quadratic exactly.

Extension

Ask students to find the rate increase the fictional district would need if the elasticity were negative 0.50 instead of negative 0.30, and to explain why a more responsive customer base makes the target harder to reach. Then have them determine the largest revenue increase the district could reach at any rate under the original elasticity, which is the maximum of the revenue function rather than a solution to a target, and discuss whether the model can be trusted at that rate.

ELL supports

Pre-teach "revenue," "elasticity," "demand response," and "viable" with a labeled two-arrow diagram showing the rate going up while average use goes down. Provide sentence frames for question 10, such as "This root cannot be used because the estimate of ___ was based on ___," and allow students to show question 7 as a labeled expansion rather than in prose.

Extensions

Students can rebuild the fictional Fetterly model with a two-tier rate, in which the first 4,000 gallons are billed at one price and everything above at a higher one, and describe why the revenue expression is no longer a single quadratic. A second follow-up asks students to find a published rate study from any nearby utility and report whether it states a demand response at all, since many do not.