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# Burning The Candle At Both Ends.....

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Two candles are the same height but have different cross sections.......one candle takes 4 hours to burn down completely, while the other takes 7 hours to burn down completely......

The question is...assuming that they both burn at a steady rate...if we light both at the same time, how long will it take for one candle to be twice as tall as the other???

CPhill  Dec 10, 2015

#3
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Two candles are the same height but have different cross sections.......one candle takes 4 hours to burn down completely, while the other takes 7 hours to burn down completely......

The question is...assuming that they both burn at a steady rate...if we light both at the same time, how long will it take for one candle to be twice as tall as the other???

H = height to begin.

h = height during the burning

Candle 1:

$$\begin{array}{rcl} h_1 = -\frac {H} {4\ \mathrm{hours}} \cdot t + H \end{array}$$

Candle 2:

$$\begin{array}{rcl} h_2 = -\frac {H} {7\ \mathrm{hours}} \cdot t + H \end{array}$$

condition:

$$\begin{array}{rcll} 2\cdot h_1 &=& h_2 \\ 2\cdot \left( -\frac {H} {4\ \mathrm{hours}} \cdot t + H \right) &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \\\\ -\frac {2\cdot H} {4\ \mathrm{hours}} \cdot t + 2\cdot H &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \qquad &| \qquad :H \\\\ -\frac {2} {4\ \mathrm{hours}} \cdot t + 2 &=& -\frac {1} {7\ \mathrm{hours}} \cdot t + 1 \\\\ -\frac {t} {2} + 2 &=& -\frac {t} {7} + 1 \\\\ -\frac {t} {2} + 1 &=& -\frac {t} {7} \\\\ \frac {t} {2}-\frac {t} {7} &=& 1 \\\\ t\left ( \frac {1} {2}-\frac {1} {7} \right) &=& 1 \\\\ t &=& \frac{1}{ \frac {1} {2}-\frac {1} {7} } \\\\ t &=& \frac{14}{ 7-2 } \\\\ t &=& \frac{14}{ 5 } \\\\ t &=& 2.8\ \mathrm{hours} \end{array}$$

$$\begin{array}{rcll} h_1 &=& -\frac {H} {4\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H \\ h_1 &=& -0.7 H + H \\ h_1 &=& 0.3H \\\\ h_2 &=& -\frac {H} {7\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H\\ h_2 &=& -0.4 H + H \\ h_2 &=& 0.6H \\\\ \dfrac{h_2}{h_1} &=& \dfrac{0.6H}{0.3H}= 2 \end{array}$$

heureka  Dec 10, 2015
#1
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Are they a different width? Forgive me if my answer makes absolutely no sense it is 3:00am over here and I just woke up.

Guest Dec 10, 2015
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They are of different widths, but.....that doesn't really factor in to the answer.....

CPhill  Dec 10, 2015
#3
+19994
+15

Two candles are the same height but have different cross sections.......one candle takes 4 hours to burn down completely, while the other takes 7 hours to burn down completely......

The question is...assuming that they both burn at a steady rate...if we light both at the same time, how long will it take for one candle to be twice as tall as the other???

H = height to begin.

h = height during the burning

Candle 1:

$$\begin{array}{rcl} h_1 = -\frac {H} {4\ \mathrm{hours}} \cdot t + H \end{array}$$

Candle 2:

$$\begin{array}{rcl} h_2 = -\frac {H} {7\ \mathrm{hours}} \cdot t + H \end{array}$$

condition:

$$\begin{array}{rcll} 2\cdot h_1 &=& h_2 \\ 2\cdot \left( -\frac {H} {4\ \mathrm{hours}} \cdot t + H \right) &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \\\\ -\frac {2\cdot H} {4\ \mathrm{hours}} \cdot t + 2\cdot H &=& -\frac {H} {7\ \mathrm{hours}} \cdot t + H \qquad &| \qquad :H \\\\ -\frac {2} {4\ \mathrm{hours}} \cdot t + 2 &=& -\frac {1} {7\ \mathrm{hours}} \cdot t + 1 \\\\ -\frac {t} {2} + 2 &=& -\frac {t} {7} + 1 \\\\ -\frac {t} {2} + 1 &=& -\frac {t} {7} \\\\ \frac {t} {2}-\frac {t} {7} &=& 1 \\\\ t\left ( \frac {1} {2}-\frac {1} {7} \right) &=& 1 \\\\ t &=& \frac{1}{ \frac {1} {2}-\frac {1} {7} } \\\\ t &=& \frac{14}{ 7-2 } \\\\ t &=& \frac{14}{ 5 } \\\\ t &=& 2.8\ \mathrm{hours} \end{array}$$

$$\begin{array}{rcll} h_1 &=& -\frac {H} {4\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H \\ h_1 &=& -0.7 H + H \\ h_1 &=& 0.3H \\\\ h_2 &=& -\frac {H} {7\ \mathrm{hours}} \cdot 2.8\ \mathrm{hours} + H\\ h_2 &=& -0.4 H + H \\ h_2 &=& 0.6H \\\\ \dfrac{h_2}{h_1} &=& \dfrac{0.6H}{0.3H}= 2 \end{array}$$

heureka  Dec 10, 2015
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Thanks Chris and Heureka,

This looks like a really interesting question :)

Melody  Dec 10, 2015
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Good job, heureka.....here is how I solved it.....

Each hour, the candle that burns down more quickly loses  7/28  of its height.....and the other candle loses 4/28 of its height.....so......we want to know when.....

2 ( h  - [7/28]h* t )  =  (h - [4/28]h* t)       where h is the original height and t is the time we are looking for, in hours........

2h - [14/28]h*t =  h - [4/28]h*t

1h =  [10/28]h * t       divide through by h

1 = [10/28] * t

28/10 = t   = 2.8 hours

CPhill  Dec 10, 2015
edited by CPhill  Dec 10, 2015