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Lesson plan Grade 8 Engineering & Design

Ironbridge Community Garden's New Pump: A Break-Even Equation Lesson

A 45-minute grade 8 lesson writing and solving a linear equation to find the break-even operating hours between two irrigation pump options.

Objectives

  1. Students will write and solve a linear equation to find the break-even number of operating hours between two pump options.
  2. Students will calculate each pump's total cost at a stated usage level and recommend a choice based on the result.
  3. Students will explain why the pump with the higher purchase price can still be the better long-term choice.

Vocabulary

Key words: break-even point, capital cost, operating cost, linear equation

Teacher overview

The fictional Ironbridge Community Garden needs a new irrigation pump, and its volunteer board is choosing between a cheaper standard pump and a pricier high-efficiency model that costs less to run. This lesson has students write a total-cost equation for each pump, solve for the number of operating hours at which both pumps cost the same, and use that break-even point to recommend which pump the garden should buy.

A common misconception in this math is judging the two pumps only by their purchase price, ignoring the fact that a linear equation with a different rate of change can cross and overtake a cheaper starting price given enough time. The procedure is built to correct that directly, since the guided and independent practice both require students to calculate total cost at more than one usage level before making a recommendation.

Procedure

Hook (5 min): Tell students that Ironbridge Community Garden's board is choosing between two pumps: Pump A costs $180 to buy, and Pump B costs $260 to buy. Ask which pump they would choose knowing only the purchase prices, and have a few students share their reasoning. Tally the class's choices on the board before revealing what each pump costs to run.

Direct instruction (10 min): Reveal that Pump A costs $0.15 in energy for every hour it runs, while the pricier Pump B costs only $0.07 per hour to run. Model writing a total-cost equation for each pump in the form C(h) = purchase price + rate × h, where h is hours of operation: C_A(h) = 180 + 0.15h and C_B(h) = 260 + 0.07h. Explain that setting the two expressions equal to each other will find the number of hours at which both pumps cost the same total amount, the break-even point.

Guided practice (15 min): As a class, set 180 + 0.15h equal to 260 + 0.07h and solve the linear equation step by step for h, with students following along and solving the same equation on their own paper. Check the answer by substituting the solved value of h back into both original equations to confirm the two costs match.

Independent practice (10 min): Tell students the garden's board estimates the pump will run about 600 hours during a single growing season. Each student calculates the total cost of Pump A and Pump B at 600 hours, then calculates the total cost of each pump over three growing seasons (1,800 hours), and writes which pump is cheaper at each of the two usage levels.

Closure / exit ticket (5 min): Students answer one question on an index card: "In one sentence, explain why the pump with the higher purchase price can still be the better long-term choice for Ironbridge Community Garden." Collect and scan for correct reasoning about operating hours before dismissal.

The quantities in this problem describe a fictional design scenario.

Ironbridge Community Garden's New Pump: A Break-Even Equation Lesson — Grade 8 Engineering & Design
Name: ____________________________ Date: ______________

Student Handout

The Ironbridge Community Garden is a fictional garden choosing between two irrigation pumps. Pump A costs $180 to buy and $0.15 in energy for every hour it runs. Pump B costs $260 to buy and $0.07 in energy for every hour it runs. Write each pump's total cost in the form C(h) = purchase price + hourly rate × h, where h is the number of operating hours.

  1. Set the two total-cost expressions equal to each other and solve for h, the break-even number of operating hours.
  2. Substitute your break-even value of h back into both total-cost equations. What total cost does each pump reach at that point?
  3. The garden's board estimates the pump will run about 600 hours in one growing season. What is Pump A's total cost at 600 hours?
  4. What is Pump B's total cost at 600 hours?
  5. At 600 hours, which pump is cheaper, and by how many dollars?
  6. Over three growing seasons the pump runs about 1,800 hours. What is Pump A's total cost at 1,800 hours?
  7. What is Pump B's total cost at 1,800 hours?
  8. At 1,800 hours, which pump is cheaper, and by how many dollars?
  9. In one sentence, explain why the pump with the higher purchase price can still be the better long-term choice for Ironbridge Community Garden.

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: 1,000 hours.
    Break-even equation: 180 + 0.15h = 260 + 0.07h. Subtract 0.07h from both sides: 180 + 0.08h = 260. Subtract 180 from both sides: 0.08h = 80. Divide both sides by 0.08: h = 1,000 hours.
  2. Answer: $330.
    Check: Pump A at 1,000 hours: 180 + 0.15(1,000) = 180 + 150 = 330. Pump B at 1,000 hours: 260 + 0.07(1,000) = 260 + 70 = 330. Both equal $330, confirming the break-even point.
  3. Answer: $270.
    600 hours (one season), Pump A: 180 + 0.15(600) = 180 + 90 = 270.
  4. Answer: $302.
    600 hours (one season), Pump B: 260 + 0.07(600) = 260 + 42 = 302.
  5. Answer: $32.
    Pump A is cheaper by $32 at 600 hours.
  6. Answer: $450.
    1,800 hours (three seasons), Pump A: 180 + 0.15(1,800) = 180 + 270 = 450.
  7. Answer: $386.
    1,800 hours (three seasons), Pump B: 260 + 0.07(1,800) = 260 + 126 = 386.
  8. Answer: $64.
    Pump B is cheaper by $64 at 1,800 hours.
  9. Answer: Pump B, once the garden runs it for more than 1,000 hours.
    Exit ticket answer: Pump B, despite its higher $260 purchase price, costs less per hour to run ($0.07 versus $0.15), so once the garden uses it for more than 1,000 hours, its lower operating cost makes up for the higher starting price and it becomes the cheaper pump overall.

Differentiation

Support

Provide both total-cost equations already written on a handout, and walk through the first two steps of solving the break-even equation as a class before releasing students to finish the remaining steps on their own. Allow a calculator throughout.

Extension

Ask students to find how many operating hours it would take for a third, hypothetical Pump C, costing $320 to buy with a $0.03 hourly operating cost, to become cheaper than Pump B, setting up and solving a new break-even equation.

ELL supports

Provide a simple diagram showing "purchase price" as a starting bar and "operating cost" as small bars added on top for each hour, so the structure of each total-cost equation is visible before students work with the symbols. Provide a sentence starter for the exit ticket: "Pump B is the better long-term choice because ___."

Every figure in this scenario is fictional, chosen so students work with realistic engineering numbers.

Extensions

Have students estimate how many hours per season their own school's water fountains, sprinklers, or another water-using device might run, and calculate what usage level would make a hypothetical higher-efficiency replacement worth its added purchase cost, using the same break-even method from this lesson.