OK, Luke, was it the first solution or the one I offered after Fred W asked what he asked?

If the first, which of the two solutions did you find easier to follow? More convincing? More elegant? Etc. ... .

I'll give Fred 1/2 a point for his help and directions.....
As for your solution Sarge, Your first one was the "beautiful" one.... but your second one convinced me more...

I've got one more also....
Q25:If a cylindrical water tank has a height of 10 and a diameter of 7 calculate the amount of sheet metal needed to create this tank. also Calculate the volume of water that this cylindrical tank will hold.

Best Regards,
Luke.

Without calculator and assuming laser cutting to minimise material loss:

The parts can be cut from sheet that is 28.99 x 10 = 289.9 Area.

The volume is 384.89.

F.

Do we have rounding errors here?
2*Pi*3.5^2+2*Pi*3.5*10 is approximately 296.8805058 (referring to surface area, the amount of material needed for construction of the tank).

As for a night's worth of sleep, I only got 5 hours. :( It took me 30-45 minutes to wind down after I stopped doing work last night. And, in 30-60 minutes I'll be doing more work.

Luke: interesting, I figured the second solution would be the one you'd consider elegant.

Sarge,

No rounding errors. The question does not state that the tank is totally enclosed (e.g. the cold water tank in my roof does not have a top).

The wall of the tank needs a sheet that is pi*7*10=219.94.
In order to have enough sheet to cut a circle with diameter 7 from the sheet then the sheet needs to be extended by 7 on its longest axis (least waste) so the sheet would have to be 28.994 * 10 to provide enough material for an open-top tank.

i.e Area = 289.94 (to 2 d.p.)

F.

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Agreed as to your approach. Still, if the tank is open at the top, yes, you have the correct result for the area of the lateral surface (approximately 219.91 cubic ... quickly erasing that and replacing it with square ... grrr ... whatever the units were), but ... .
With only one circle, at the bottom, Pi*r^2 approximately 3.14 * (7/2)^2 = 3.14 * 3.5^2 = 3.14 * 12.25 = 38.465.
Adding the two, we're at approximately 258.375, where as your answer was very close to mine, where I made the assumption of both the bottom and top enclosed by circles.

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Capitalize on this good fortune, one word can bring you round ... changes.

Agreed as to your approach. Still, if the tank is open at the top, yes, you have the correct result for the area of the lateral surface (approximately 219.91 cubic ... quickly erasing that and replacing it with square ... grrr ... whatever the units were), but ... .
With only one circle, at the bottom, Pi*r^2 approximately 3.14 * (7/2)^2 = 3.14 * 3.5^2 = 3.14 * 12.25 = 38.465.
Adding the two, we're at approximately 258.375, where as your answer was very close to mine, where I made the assumption of both the bottom and top enclosed by circles.

I think we're at cross purposes a little.

You are calculating the surface area of the tank. I am calculating the area of sheet metal that the components can be cut from. What you are saying is that the scrap left over after I have cut my one circle is almost enough area (though entirely the wrong shape) for a second circle.

F.

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Ahhh, but it is metal and so perhaps it could be melted down and reshaped. :)
Also, unless told otherwise, a mathematical problem assumes ideal situations, such as no material loss.

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Capitalize on this good fortune, one word can bring you round ... changes.

Ahhh, but it is metal and so perhaps it could be melted down and reshaped. :)
Also, unless told otherwise, a mathematical problem assumes ideal situations, such as no material loss.

Yeah. But the question asks for the "amount of sheet metal". If I buy an amount of sheet metal I get charged for any offcuts too (and probably for the cutting). In my professional life this comes under the heading of "inadequate specification" ;)

Judgement of Solomon would be "Shoot the messenger"? Guess we have to wait for Solomon - er... Luke.

F.

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I still say, it's metal, melt it! Or, use Luke's lightsaber. ;)

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Capitalize on this good fortune, one word can bring you round ... changes.

Solve for "Planet X." :)

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Capitalize on this good fortune, one word can bring you round ... changes.

There are two winners here...
Fred wins 1 Point for the Volume Answer and....
Sarge wins 1 Point for the first correct Surface Area answer...

Correct Answer:
Q25 Volume: 384.9 Cubic Units
Working:
2Ï€rh + 2Ï€r^2
= 2Ï€(35 + 49/4)
= 189/2 π ≈ 296.9

Q25 Surface: 296.9 Square Units
Working:
Ï€r^2h
= π(49/4)(10)
= 245/2 π ≈ 384.9

And The Updated Standings:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 5 1/2 Points
3. WinterKnight - 5 Points
4. Sarge - 4 Points
5. Dominique - 3 1/2 Points
6. Fred W - 1 1/2 Points
7. John McLeod VII - 1 Point
8. Scary Capitalist - 1/2 Point
9. TBD......

Sarge and Fred quickly make a big impact to the Standings Table... doing we have a few new leaders blossoming?....

Also, My lightsaber is all powerful.... It has advanced mathematics technology..
And its Blue....

Q26, Q27 coming soon..

May the Force be with you....
Luke.

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- Luke.

Here are 3 more questions for everyone.......

Q26: There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four?

Q27:Given a $100,000 mortgage, a payment period of 30 years, monthly payments (the first due at the end of the first month), and an interest rate of 7.5% compounded monthly, what will be the monthly payments?

Q28: Thirteen pirates put their treasure in a safe.
They decide that the safe should be able to be opened if any majority of pirates agree but not be able to be opened if any minority agree.
The pirates don't trust each other so they consult a locksmith.
The locksmith puts a specific number of locks on the safe such that every lock must be opened to open the safe. Then he distributes keys to the pirates such that every pirate has some but not all of the keys. Any given lock can have multiple keys but any given key can only open one lock. What is the least number of locks required?

Best Regards,
Luke.

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- Luke.

The mortgage problem solution is $843.86 per month with the rate divided by 12 and amortization set up for 360 periods.

Thoughts on the pirate problem:

13 pirates so a majority is seven
the locksmith must make a lock for each distinct group of 6.
So you can select 6 from 13 the following number of ways:

(13x12x11x10x9x8) / (1x2x3x4x5x6) = 1716 locks !!

I must be nuts or a little rusty on combinatorics but it seems that this is the only way that any 7 but not any 6 could open the chest.

Bill

The mortgage problem solution is $843.86 per month with the rate divided by 12 and amortization set up for 360 periods.

r = 0.075
m = 12
t = 30

i = r / m = -.00625
n = m * t = 360

Periodic payments = 100000 / ((1 - (1 + i)^(-1 * n)) / i) = 699.21

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Capitalize on this good fortune, one word can bring you round ... changes.

Another try--my math is a bit sloppy: Thats why I like to use a program or EXCEL.

sum of (1+.075/12)^N for n=1 to 360 times the 100,000 all divided by 360 to make the same monthly payments and the answer is 699.21

Sarge is right my answer was for a 15 year loan and was wrong anyway.

Regards,

Bill

Another try--my math is a bit sloppy: Thats why I like to use a program or EXCEL.

sum of (1+.075/12)^N for n=1 to 360 times the 100,000 all divided by 360 to make the same monthly payments and the answer is 699.21

Sarge is right my answer was for a 15 year loan and was wrong anyway.

Regards,

Bill

Bill, I just taught the material recently, so I figured I better take a shot at it or else I'd have a heck of time scoring the final exam! I think you had the right approach.

On the pirates one, no clue yet. On the number of cereal boxes, I am trying to turn myself from what originally came to mind because I don't think I was applying the correct concept at first.

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Capitalize on this good fortune, one word can bring you round ... changes.

Sarge--I need to give this more rigor instead of the intuitive stuff I like which is usually wrong. On the problem of the 4 colors. I think about the following.

The chance (probability) of having an acceptable draw at trial 1 is 1.00
And that having been done the chance of having a successful draw on trial 2 is 3/4 or .75 and that having been done the chance of having a successful draw on the third trial is .5 since only half of the color choices would be successful. On the fourth trial the probability of success is .25
.
Therefore .75 x .5 x .25 =.0975 Taking the inverse of this is 1/.0975 =10.25 so I say that the expected number of trials would be 11.

Of course I seem to remember something about conditional probability that is missing here in my analysis. I don't even know if Bayes theory needs to be applied here --I think not


Anyhow: My answer is 11 trials

I am going to have to think about that one more, Bill.
I just read your approach quickly.
On the second draw, would you not have to consider both the situation of picking the same thing you did the first time versus picking one of the 3 remaining colors?

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Capitalize on this good fortune, one word can bring you round ... changes.

Another try--my math is a bit sloppy: Thats why I like to use a program or EXCEL.

sum of (1+.075/12)^N for n=1 to 360 times the 100,000 all divided by 360 to make the same monthly payments and the answer is 699.21

Sarge is right my answer was for a 15 year loan and was wrong anyway.

Regards,

Bill

Bill, I just taught the material recently, so I figured I better take a shot at it or else I'd have a heck of time scoring the final exam! I think you had the right approach.

On the pirates one, no clue yet. On the number of cereal boxes, I am trying to turn myself from what originally came to mind because I don't think I was applying the correct concept at first.

Guess what I am going to be teaching statistics here in a grad school starting this summer--How do you spell Chebyschev ??? The only good thing is that as I relearn each concept i can figure out what the difficult ideas are and steer the students to grasping these particular ideas.

I have had many experiences where I get a refresher, or am offered up challenges, that serve me well a few months later when teaching.

I've seen two other spellings of Chebyshev, I believe.
http://en.wikipedia.org/wiki/Pafnuty_Chebyshev

His name is transliterated variously as Chebychev, Chebyshov, Tchebycheff or Tschebyscheff (French and German transcriptions) among many others.

One of the issues is the use of a different alphabet. Also, there are slight differences in what syllables a child learns to pronounce, around the world. But, I do not know if this is the case for Russian vs. English speakers.

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Capitalize on this good fortune, one word can bring you round ... changes.

P.S. - don't completely throw out the intuitive stuff. :)

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Capitalize on this good fortune, one word can bring you round ... changes.

Congratulations to Bill & Sarge.
Bill was right on Question 28: Answer is 1716 locks. 1 Point
Sarge was right on Question 27: Answer is $699.21. 1 Point

Well Done both of you!!!

And The Updated Standings:
1. Mr. Kevvy - 7 Points
2. William Rothamel - 6 1/2 Points
3. WinterKnight - 5 Points
4. Sarge - 5 Points
5. Dominique - 3 1/2 Points
6. Fred W - 1 1/2 Points
7. John McLeod VII - 1 Point
8. Scary Capitalist - 1/2 Point
9. TBD......

Bill is just 1/2 a point away from taking the title of Best Problem Solver....
Sarge is making a big impact on the table as well.

Q26 is still here:
Q26: There is a free gift in my breakfast cereal. The manufacturers say that the gift comes in four different colors, and encourage one to collect all four (thus eating lots of cereal). Assuming there is an equal chance of getting any one of the colors, what is the expected number of boxes I must consume to get all four?

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- Luke.