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September 2nd, 2018, 07:18 AM   #1
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Calculating the life (in time) of a marker by knowing the life in meters.

So, recently I wanted to buy a whiteboard marker but I also wanted to know the life of that marker to see how much time it will have until it dries. So I asked the seller to tell me and he told me that the life of the marker is 500 m.

I'm intended to use this marker on a whiteboard with dimensions 60cmX80cm

So I created this problem:

How much time will the marker last if every day I'm covering the entire board only once?

Then I solved the problem like this:

$\displaystyle
E = \frac{60cm}{100} \cdot \frac{80cm}{100} = 0.6m \cdot 0.8m = 0.48 m^{2}
$

So if I'm filling the entire board once per day then:
$\displaystyle
usePerDay = 0.48 \cdot \frac{m^{2}}{day}
$

$\displaystyle
life = \frac{totalLife}{usePerDay} = \frac{500m}{0.48 \cdot \frac{m^{2}}{day}}
= \frac{500m \cdot days}{0.48 \cdot m^{2}}
= \frac{500days}{0.48m}
= 1041.6 \cdot \frac{days}{m}
$

And as you can see, I suck creating and solving math problems.
I can't even understand what I just calculated there. What is days/meters lol.
I wanted to find just days...

probably the error here is that I used m^2 in the division and somehow I should make E in meters
let's say by unpacking somehow E and making it a straight line (graphically) that represents the same quantity but in meters? So then the result would be in days only.

Can someone help me?
Thanks.

Last edited by skipjack; September 2nd, 2018 at 07:36 AM.
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September 2nd, 2018, 07:48 AM   #2
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Why would you cover the entire board? What is the "average width" of a line drawn with the marker?
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September 2nd, 2018, 08:10 AM   #3
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Quote:
Originally Posted by skipjack View Post
Why would you cover the entire board? What is the "average width" of a line drawn with the marker?
Because I want to cover the worst case where I'm using the marker too much to cover the entire board.
Imagine that I'm painting the board trying to turn it from white to black.

Last edited by babaliaris; September 2nd, 2018 at 08:13 AM.
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September 2nd, 2018, 08:15 AM   #4
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What is your answer in regard to my second question?
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September 2nd, 2018, 08:34 AM   #5
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Quote:
Originally Posted by skipjack View Post
What is your answer in regard to my second question?
0.5cm

I calculated it based on the width of each of my letters.
Approximately the width of each letter is 0.5cm .

But still I can't understand where you are going. If you have to consider the
dimension of each letter, you also need an average height. In other words, the problem becomes a random problem because the length of each letter is actually random (you don't make the same size every time). So now we must solve the problem using probabilities...

Last edited by skipjack; September 2nd, 2018 at 09:49 AM.
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September 2nd, 2018, 10:09 AM   #6
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The so-called life of the marker refers to when it draws a line of "average" width 0.5 cm and length 500 m without self-intersection. The area marked is then 2.5 m². As your board has area 0.48 m² and is filled once per day, the marker's life is (2.5/0.48) days = 5 5/24 days.

Note that I took the width of the line to be 0.5 cm for the purpose of the above example, whereas you had chosen to give me the width of a letter, even though the marker is, in the worst case scenario, marking the entire board, and so it can't be assumed it's being used for writing letters.
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September 2nd, 2018, 10:18 AM   #7
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Can you show me how you calculated the marked area? Also in the last answer is that a 55/24?

Offtopic:
I tried to quote your answer and I got this message:
You must return when the moon has friends and the fox is borrowed.

Is this a joke?

Last edited by skipjack; September 2nd, 2018 at 10:47 AM.
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September 2nd, 2018, 10:45 AM   #8
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When I evaluated 0.5 cm × 500 m, I used the facts that 0.5 cm = 0.005 m and 0.005 × 500 = 2.5.

I intentionally wrote 5 5/24 (a mixed fraction equal to 5 + 5/24).

The message you saw was because you quoted a post that no longer existed (I had deleted and reposted my reply).
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September 2nd, 2018, 11:53 AM   #9
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Quote:
Originally Posted by skipjack View Post
When I evaluated 0.5 cm × 500 m, I used the facts that 0.5 cm = 0.005 m and 0.005 × 500 = 2.5.

I intentionally wrote 5 5/24 (a mixed fraction equal to 5 + 5/24).

The message you saw was because you quoted a post that no longer existed (I had deleted and reposted my reply).
Thanks! So the problem was lack of information? How did you know that information about the average line? Probably you have experience in a lot of real problems.

Also what I calculated is wrong? Or I found a different answer of the one I was intended to. I divided length/area , which I'm a little confused of what it means.

Last edited by babaliaris; September 2nd, 2018 at 11:58 AM.
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September 3rd, 2018, 12:04 AM   #10
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Markers vary considerably, so a typical line would probably be a lot narrower than 0.5 cm.

The marker's lifetime will presumably depend on the line's width as well as its length, so your calculation (that effectively ignored the width) was inappropriate and dimensionally incorrect.
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