Calling Math Peoples! Fun inside!

[Jotaru]

Silv > I get what you're saying, but I'm trying to find Y's in respect to X where X will be supplied, not the other way around. Ideally yeah I'd make X infinitesimal and increase their quantity to infinity, but that isn't what I'm doing; i'm going to be provided a certain X and then go from there.

The second problem actually has two variables, X and quantity, so I'm going to just not bother with it, lol.

Edit: And thank you for solution to problem A

This is all MMO related in an odd way... Promise.

Zhais around?

[netz]

Is your second problem just trying to pack as many spheres as you can inside a larger sphere?

Or is it an extension of your first problem where what you're attempting to do is something like this:

http://www.zten.org/img/circles.png

Except with as many circles as possible, centered a distance from the larger circle's center so that the surface area of the larger enclosed in the smaller circle is maximized?

[Jotaru]

NVM, #2 is easily solvable, I was just being a retard when I wrote it.

[Zhais]

Silv's answer is exactly what jotaru is looking for I think... I don't really get why he's trying to do this, but he is >_> (Are you trying to put a bubble up over a bubble? 0.o)

[Jotaru]

No, trying to optimize bubbles on a gate.

Tin thinks placing people 15km from gate and orbitting is best and I was convinced it wasnt

Turns out I was right!

[Zhais]

Nowwwww I get it...

yea, want bubbles just inside that 15km range, or want to station people around the same area

[Jotaru]

And that formula tells us where to tell people to orbit, X being their Scram range

[Zhais]

Yay maths!

o/ \o

[Trajan]

Do you want to keep this thread going with more problems? If so here is one I worked on but was unable to fully solve....

Let x be the solution of x(3^x) = 3^18. Find an integer k such that k < x < k+1.

The answer is obvious if you plug it into your calculator, but I am interested in showing it w/out using one.
It seems pretty clear that x =/= 1 and x =/=18, so x is bigger than 1 and less than 18. So k is one of 1,2,3,...,16,17.
The best I have been able to do is show that k = 15,16 or 17. So yea, Kudos to whoever can show what k is exactly equal to.

[Silviya]

There is probably a better way of proving it using logarithms, but just using basic math and proof by induction.

If x = 18, then 18(3^18) which is > 3^18
If x = 17, then 17(3^17) which is < 3^18
Thus, 17 < x < 18

So k = 17 such that k < x < k+1

[Suiram]

There is probably a better way of proving it using logarithms, but just using basic math and proof by induction.

If x = 18, then 18(3^18) which is > 3^18
If x = 17, then 17(3^17) which is < 3^18
Thus, 17 < x < 18

So k = 17 such that k < x < k+1

Except that 17(3^17) > 12(3^17) = 4*3(3^17) = 4(3^18) > 3^18


On the other hand, 15(3^15) < 27(3^15) = 3^18, while 16(3^16) > 9(3^16) = 3^18, so k=15

[Zanmato]

As a preface, no, this isn't homework, its actually something recreational... seems odd perhaps, but yeah. Problems like this eat at me and its been forever since I've been in geometry so I figured come here first, in case anyone can spew out some numbers offhand.

I'm basically looking for a few different things. To set it up: You have a sphere of radius 15km centered on point A. You want to set up a sphere with radius X to cover the maximum amount of surface area on the 15km sphere... trying to figure out at what distance you would locate (from point A) to maximize surface area coverage.

As a secondary problem, looking to maximize internal coverage volume assuming multiple X km spheres... not sure how best to word this. The first logical thing that springs to mind is just to place a sphere at the center- but, obviously, if you have, say, 3 of these smaller spheres, you can't place them all at A.

Bonus points if you can figure out what this is for

BTW, you don't necessarily need to work it out for me if you don't feel, but if you have general formulas / tips to get in the right direction, that's appreciated too!


Aaaaaand go

First, let me answer the above question.

Bonus points if you can figure out what this is for

Is this an optimization problem? Do we have a winrar?!?!

Anyway ...

There is something not right with this question. The problem is that x is not fixed and because of that, let us see what happens:

Lets start off solving this problem how we would solve any optimization problem. (Pretend there is nothing wrong with this question for now.)

Mr. Pythagoras and his famous Pythagorean Theorem is your friend here.

Clearly, y = (( 15^2 ) - ( x^2 ))^0.5
or
y = sqrt(( 15^2 ) - ( x^2 )).
(I like the first one since I am going to apply the Chain Rule to it.)

And now you have to find out the maximum value of y such that the above equation holds. Easy. Gotta find dy/dx.

y = (( 15^2 ) - ( x^2 ))^0.5
Apply Chain Rule and you get …
dy/dx = 0.5[(( 15^2 ) - ( x^2 ))^-0.5]*(-2x) = -x[(( 15^2 ) - ( x^2 ))^-0.5] = -x/y

Now to find the maximum value of y, you have to set dy/dx = 0.
So –x/y = 0
Which gives x = 0.

This is where you should realize that something has gone wrong. x cannot equal to zero. But in this case, x is in fact equal to zero. Let me explain why.

(In the ideal optimization problems, when you set dy/dx = 0, x equals to some real non-zero value. You would still follow these steps though.)

Ok, so let me explain why x = 0 is correct here.

Let us see what happens if we plug in x = 0 in y.
y = (( 15^2 ) - ( 0^2 ))^0.5 = 15.

Therefore the answer to this question is 15. You might think that this is wrong since 15 is actually the radius of Sphere A, but if you think about it mathematically, the only way for the smaller sphere to cover the maximum surface area of the bigger sphere is if they have the same radius. You could think of it like they’re overlapping each other.

And that is why earlier I said there is a problem with letting x vary and not making it a fixed value. The Math will just force the 2 radii to be equal.

Now, if you are a Non-Mathie and the above didn’t make sense (I’m not sure if you are familiar with the Chain Rule), here is a logical explanation:

We agree that the following equation is correct, amirite? It is simply the Pythagorean Theorem:

y = (( 15^2 ) - ( x^2 ))^0.5
or
y = ( 225 - x^2 )^0.5
(Obviously 0 <= x <= 225 since you can’t find the square root of a negative number.)

Toss your calculator/pen/paper in somewhere and just stare at this for a good 10 minutes. Using some logic, try to figure out what value of x will give you the biggest/maximum value for y. (Which is what we are trying to figure out, i.e. the value of y).

y = ( 225 - x^2 )^0.5

You will realize that no matter what value of x you choose, you’re squaring it [the x^2 part in y = ( 225 - x^2 )^0.5] which will make it a big positive number but you want maximum y and that will only happen when x = 0.

Like I said x = 0 doesn’t make sense because in reality if you try to make two balls with clay you cant make a ball with radius equal to 0 but mathematically speaking it is right because this would give y = 15 which would mean the two balls have to have the same radius and would therefore overlap each other for maximum surface area.

[Zanmato]

Do you want to keep this thread going with more problems? If so here is one I worked on but was unable to fully solve....

Let x be the solution of x(3^x) = 3^18. Find an integer k such that k < x < k+1.

The answer is obvious if you plug it into your calculator, but I am interested in showing it w/out using one.
It seems pretty clear that x =/= 1 and x =/=18, so x is bigger than 1 and less than 18. So k is one of 1,2,3,...,16,17.
The best I have been able to do is show that k = 15,16 or 17. So yea, Kudos to whoever can show what k is exactly equal to.

The “without using a calculator” part is what made me want to solve this.

First, I want to make sure there is only one solution to x(3^x) = 3^18. If we can show that it does really have only one solution, we can plug in numbers to guess a value of x and we’ll know that it’s the only value and the right one.

Clearly, x has to be > 0. So we can use logarithms.

x(3^x) = 3^18
ln[x(3^x)] = ln[3^18]
ln[x]+ln[(3^x)] = ln[3^18]
ln[x]+xln[3] = 18ln[3]
ln[x] = - xln[3] + 18ln[3]

Let y = ln[x]
Therefore y = - xln[3] + 18ln[3] = (-ln[3])x + 18ln[3]

So let’s check out how these 2 lines would look on a graph.

y = ln[x]
Basic graph. No turning points.

y = (-ln[3])x + 18ln[3]
In the form y = mx + b. Straight line.

Ok, so now we know these two lines intersect for sure and most importantly at only one place, so there is only one solution to x(3^x) = 3^18 and it is real.

The above was just to make sure that solution does in fact exist (as opposed to a curve and a line passing close to each other but not intersecting).

Back to answering the question:

x(3^x) = 3^18.
x = 3^(18-x)

LHS = x
RHS = 3^(18-x)

It seems pretty clear that x =/= 1 and x =/=18, so x is bigger than 1 and less than 18. So k is one of 1,2,3,...,16,17

Skipped x = 1 to 11 in table (felt lazy, sorry >.<)

x……..LHS……RHS…………….....|RHS – LHS|
12…..12……..3^6 = 729……...717
13…..13………3^5 = 243……...230
14…..14………3^4 = 81………..67
15…..15………3^3 = 27………..12
16…..16………3^2 = 9………....7
17…..17………3^1 = 3………....14

I apologize for my lame ass table. Pretend the dots are invisible.

Note:

RHS is the Right-Hand Side.
LHS is the Left-Hand Side.
|RHS – LHS| is the modulus of the difference of the Right-Hand Side and the Left-Hand Side.

Therefore, I conclude that 15 < x < 16.
So k = 15.

[Trajan]

Yep k is equal to 15, GJ you two. Seems so obvious now! Why don't we have a person, the 1st person to solve the previous problem usually, post a new problem, and then whoever solves that problem first can post a new problem, etc etc to keep the thread going?