Lesson goal: The two burning candles and linear rates of change

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In the 1967 MAA High School Mathematics Contest (Problem #27), contestants faced this classic algebra challenge:

"Two candles of the same length are made of different materials so that one burns out completely at a uniform rate in 3 hours and the other in 4 hours. At what time should the two candles be lighted simultaneously so that, at 4:00 PM, one stub is twice the length of the other?"

Let each candle have an initial length of $L_0 = 1$ (representing 100% of its height).

Because both candles burn down at a constant, uniform speed:
  • Candle A: Burns completely in 3 hours. Its burn rate is $r_A = \frac{1}{3}$ length/hour. $$L_A(t) = 1 - \frac{t}{3}$$
  • Candle B: Burns completely in 4 hours. Its burn rate is $r_B = \frac{1}{4}$ length/hour. $$L_B(t) = 1 - \frac{t}{4}$$
Since Candle A burns faster, Candle B will be the longer stub. We seek the elapsed burning time $t$ where Candle B's height is double Candle A's height: $$L_B(t) = 2 \cdot L_A(t)$$ $$1 - \frac{t}{4} = 2\left(1 - \frac{t}{3}\right) = 2 - \frac{2t}{3}$$ Rearranging the terms: $$\frac{2t}{3} - \frac{t}{4} = 2 - 1 = 1$$ Find a common denominator ($12$): $$\frac{8t - 3t}{12} = 1 \implies \frac{5t}{12} = 1 \implies t = \frac{12}{5} = 2.4\text{ hours}$$ Converting $2.4$ hours into hours and minutes: $$0.4\text{ hours} \times 60\text{ minutes/hour} = 24\text{ minutes} \implies t = 2\text{ hours and } 24\text{ minutes}$$ Counting backwards from 4:00 PM: $$4:00\text{ PM} - 2\text{ hours } 24\text{ minutes} = 1:36\text{ PM}$$ In this lesson, we write code to simulate the burn down process and find the exact lighting time!
rate_A = 1 / 3
rate_B = 1 / 4
for t = 0, 3, 0.1 do
  len_A = 1 - rate_A * t
  len_B = 1 - rate_B * t
  if math.abs(len_B - 2 * len_A) < 0.05 then
    print(t)
  end
end
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  • Linear decay: Both functions $L_A(t)$ and $L_B(t)$ are straight lines with negative slopes ($-1/3$ and $-1/4$). The steeper line drops faster.
  • Verifying the heights: At $t = 2.4$ hours:
    • Candle A height: $1 - 2.4 / 3 = 1 - 0.8 = 0.20$
    • Candle B height: $1 - 2.4 / 4 = 1 - 0.6 = 0.40$
    • $0.40 = 2 \times 0.20$ — exactly double!

Now you try. Run the code to verify that both candle stubs have a 2:1 ratio at 1:36 PM. Then see Example 2 to plot the burning candles on the screen!

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