Question

A candle is 7 in tall after burning for 1 h. The same candle is 5.5 in tall after burning for 4 h. How tall will the candle be after burning for 6 h?

Answer 1

`l = (15 - t)/2` hours
The length of the candle (`l`) after 6 hours will be 4.5 inches.

Explanation:

graph{x = (15 - y)/2 [-1, 10, -4.56, 16]}

We know that the length of the candle changed from 7 inches to 5.5 inches between time 1 hour and 4 hours. The rate of change can be calculated in inches per hour.

`m = (Delta h)/(Delta t) = (h_f - h_i)/(t_f - t_i)`
`m = (5.5-7)/(4 - 1) = 1.5/3 = -1/2`

The height at time 0 can be calculated by plugging in the slope `m` into our line equation:
`y = mx + b`
We know `y` at time 4 hours is 7 inches.
`7 = -1/2 * 1 + b`
`7 = -1/2 + b`
`7.5 = b`

Plugging both of those values into the equation and using the variables `l` for the length of the candle and `t` for time we get:
`l = -1/2 * t + 7.5`

And we can calculate that at time `t=6` the length of the candle will be:
`l = -1/2 * 6 + 7.5`
`l = -3 + 7.5`
`l = 4.5`

Here's a video of a candle burning over a couple of hours to inspire you:
Here's a video

Answer 2

`color(red)("There is a trap in this question")`

After 6 hours the height will be 4.5 inches

Explanation:

The time to reduce to the 7 inches tall is of importance. Unfortunately we do not have the starting height.

`color(blue)("Determine the ratio of the burning rate")`

`("distance B to C")/("time B to C")->(7-5.5)/(4-1)color(magenta)(=1.5/3" as a ratio")`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine the measurement B to D "-> x)`

`("distance B to D")/("time B to D")-> x/(6-1) = x/5`

This will have the same ratio as our previously calculated burning rate giving:

`1.5/3-=x/5` where `-=` means equivalent to.

Matching denominators: to change 3 to 5 we multiply by `5/3`

So if we multiply the denominator by `5/3` we have to do the same to the numerator.

`color(green)(1.5/3color(red)(xx1) " "->" "1.5/3color(red)(xx(color(white)(.)5/3color(white)(.))/(5/3))" "->" "2.5/5=("distance B to D")/("time B to D")`

`color(magenta)(x/5=2.5/5=> x=2.5)`

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Determine distance D to E")`

`h=DE" "=" "BE-x" " =" " 7-2.5 = 4.5`