1. ## Solve 0=n(8pi^3-20pi^2*v+16pi^2-3*v^3)+p(8pi^2*v-3*v^3)

Hi!
Is there any way to solve this equation? If there is, can you explain it to me step by step? If not, why? (The question is: which values for v satisfies the equation?)

0=n(8pi^3-20pi^2*v+16pi^2-3*v^3)+p(8pi^2*v-3*v^3)
Where n=sqrt(4pi^2-v^2)
and p=sqrt(4pi*v-v^2)

Thank you!
-TheHurricane

2. Originally Posted by TheHurricane
Hi!
Is there any way to solve this equation? If there is, can you explain it to me step by step? If not, why? (The question is: which values for v satisfies the equation?)

0=n(8pi^3-20pi^2*v+16pi^2-3*v^3)+p(8pi^2*v-3*v^3)
Where n=sqrt(4pi^2-v^2)
and p=sqrt(4pi*v-v^2)

Thank you!
-TheHurricane
Start by substituting the values of $n$ and $p$ and simplify as much as possible.

3. That equation sure looks "suspicious"; where does it come from?

4. @Prove it; That's exactly what I've done. N and P are substitutions and I have simplified as much as I can already. Now I'm stuck though...

@Wilmer; The problem is as follows:
* Cut off a sector with the angle v.
* Now create cones of both the remaining sector and the sector cut out.
* Determine the angle v so that the total volume of the two cones are as large as possible.

I get my equation after deriving the equation for the total volume, and then putting it equal to zero (to find maxpoints). This is where I'm getting stuck.

Volume for "big" cone:

Volume for "small" cone:

Total volume for both cones:

I'm from Sweden so my english isn't perfect, but I hope you understand what I'm trying to say. Help is appreciated!

- TheHurricane

5. Can't access your 3 links (li'l square with a red "x"!),
but I think I get what you mean: is there an answer given?
I get v = 117 degrees (closest integer) doing it quickly by brute force;
the 2 volumes = .45664.... (using circle radius 1, of course)

NOTE:
You show this:
> Where n = sqrt(4pi^2 - v^2)

Since 4pi^2 = ~39.478 then (since v^2 < 4pi^2) maximum v = sqrt(39.478) < ~6.283 ;
Do you see what I mean? That's impossible. HOW in heck did you get that?

6. ## I suspect chain rule

I attacked the problem, and here's what I can tell you:

*You WILL be stuck with a polynomial of degree at least 4. It's messy. It may even be 5. I'd be very impressed to see this done without a calculator.

*You missed one shortcut: you could have COMPLETELY eliminated pi from your equation by solving for the radius of the small cone. We have Volume = 1/3 pi r^2 h, where the only occurrence of pi in the whole equation is the 1/3 pi at the beginning. You can use the chain rule as follows:

-V is the TOTAL VOLUME
-r is the RADIUS of the SMALL CONE
-o is THETA, the angle you cut

we know we want dV/do = 0

dV/do = dV/dr * dr/do = 0. Now, since dr/do = 1/(2pi) is constant, therefore never 0, we divide both sides by it, and have

dV/dr = 0. Now we simply factor out the 1/3 pi from the volume equation, and we are now solving for r rather than theta. And not a pi in sight.

The equation still involves square roots, however, due to the unfortunate Pythagorean formula involved in finding the height of the cones. Unless there are some miraculous cancellations, eliminating the two radicals will give you a fifth degree polynomial to solve. Get out your calculator.
================================================== =====
Now, I don't think that you're a multi-variable student, so multi-variable methods may not be at your disposal. However, there is a method called a "Lagrange Multiplier." I used it to QUICKLY PROVE that the cones MUST BE THE SAME SIZE (theta = pi).

The Lagrange Multiplier method would work as follows:

$V = f(x,y) = \cfrac{\pi}{3}(x^2\sqrt{1-x^2}+y^2\sqrt{1-y^2})$

Don't forget, x is the small cone's radius, and y is the large cone's radius. Now, for simplicity, let's assume that the original circle (from which we're making the cones) has a radius of 1. Then it's easily shown that x + y = 1. This is called a constraint equation, g(x,y), because it constrains or restricts the values we can choose for x and y.

$g(x,y) = x+y = 1$

Lagrange figured out that f(x,y) will only be maximized when the gradient of f(x,y) is equal to some constant $\lambda$ times the gradient of g(x,y).

$\bigtriangledown V = \bigtriangledown f(x,y) = \lambda \bigtriangledown g(x,y)$

This, as it turns out, is a wonderfully easy equation to solve:

$\cfrac{-x}{\sqrt{1-x^2}} = \lambda$

$\cfrac{-y}{\sqrt{1-y^2}} = \lambda$

thus, we have

$\cfrac{-x}{\sqrt{1-x^2}} = \cfrac{-y}{\sqrt{1-y^2}}$

which can only occur when

$x = y$

You don't need to understand Lagrange Multipliers to know that this is a simpler, more beautiful, more symmetric way of solving problems.

In my opinion, multi-variable can and should be taught along-side single variable calculus. Often, as in this problem, single-variable calculus is an ugly and unnatural way of solving problems.

--Jake

7. I get this as the 2 volumes:

SQRT[1 - a^2][pi(a^2)] where a = v/360
and
SQRT[1 - b^2][pi(b^2)] where b = (360 - k)/360

Seems to work out ok.
But how to determine the MAX from that is beyond me

8. That's easy: Add the equations for a and b together:

$\cfrac{v}{360}+\cfrac{360-v}{360}=1$

That's how I derived my equation, g(x,y) = x+y = 1 (I call them x and y, you call them a and b).

Then, if you don't know how to find the gradient of a function ( $\bigtriangledown f(x,y)$), I would suggest looking it up on Wikipedia or google or whatever. It's really simple.

how to determine the MAX from that is beyond me
The equation is

$\bigtriangledown f(x,y) = \lambda \bigtriangledown g(x,y)$

Where $\lambda$ is some (unknown) number. $\lambda$ in this equation is called the Lagrange Multiplier.

9. Originally Posted by Mazerakham
I used it to QUICKLY PROVE that the cones MUST BE THE SAME SIZE (theta = pi).
What do you mean, "same size"? Cut the circle in half?

10. @Wilmer: Yeah I get the answers v1=117 degrees, v2=pi (min-point though) and v3=2pi-v1=243 degrees. But that's with computer help; I want to know how to do it the "right way". These are the equations I got which didn't show in my last post:
Volume for "big" cone: (2pi-v)^2/(24*pi^2)*(sqrt(4pi*v-v^2))
Volume for "small" cone: v^2/(24pi^2)*(sqrt(4pi^2-v^2))
Total volume for both cones: (2pi-v)^2/(24*pi^2)*(sqrt(4pi*v-v^2))+v^2/(24pi^2)*(sqrt(4pi^2-v^2))
This is what the function looks like:

And I have no idea how I ended up with that impossibility , sorry. Maybe I should post all my calculations but that would take some time...

@Mazerakham: Do you mean I can solve the problem by finding which r gives the largest volume (instead of which v)?

And v=pi (which i think x=y means in your equation?) gives the min-point in the plot above. How about the two max-points? If you can explain that a little further I might be tempted to learn your method.

Also, since my plot got three min-/maxpoints; doesn't that mean it's a third-degree polynomial?

@Mazerakham (2): That equation works for all v's...?

Thanks for help all! I'll try to do the equation with r instead of v. Let's see how it goes.
- TheHurricane

11. Okey, now I've tried with r instead but I end up with the same problem...Now the equation looks like this:
0=n*pi*r*(2-3r^2)+p*pi(1-r)(1-6r+3r^2)
Where n=sqrt(2r-r^2)
p=sqrt(1-r^2)

Was a good idea though
I got like 1-2 more days to solve it.

12. Originally Posted by TheHurricane
@Wilmer: Yeah I get the answers v1=117 degrees, ..........
Volume for "small" cone: v^2/(24pi^2)*(sqrt(4pi^2-v^2))
BUT sqrt(4pi^2 - v^2) = sqrt(4pi^2 - 117^2) : IMPOSSIBLE.
What am I missing?

Btw, 116.6378....

13. ## I apologize

As excited as I was, it turns out I only gave you guys a minimum. It also turns out that I calculated the gradient wrong, which is embarrassing.

The gradient, calculated correctly, with pi / 3 factored out is:

$<2x\sqrt{1-x^2}\ -\ \cfrac{x^3}{\sqrt{1-x^2}}\ \ ,\ \ 2y\sqrt{1-y^2}\ -\ \cfrac{y^3}{\sqrt{1-y^2}}>$.

(This is a vector, <A,B>, in case my notation confuses you).

Now, we'll find that the Lagrange Multiplier method will give us the same equation. We have

$<2x\sqrt{1-x^2}\ -\ \cfrac{x^3}{\sqrt{1-x^2}}\ \ ,\ \ 2y\sqrt{1-y^2}\ -\ \cfrac{y^3}{\sqrt{1-y^2}}>\ =\ \lambda\ <1,1>$

Thus, equating the x and y components of these vectors, we have the egregious equation:

$2x\sqrt{1-x^2}\ -\ \cfrac{x^3}{\sqrt{1-x^2}}\ =\ 2(x-1)\sqrt{2x-x^2}\ -\ \cfrac{(x-1)^3}{\sqrt{2x-x^2}}$

Now, we see that we have TWO DIFFERENT radicals. First, we have to rationalize the denominator by multiplying through by BOTH of these radicals.

After this step, you'll see that you no longer have anything in the denominator. However, you'll be left with the two radicals in the numerator.

To get rid of these, you would multiply by the "conjugate". This step will leave you with an EIGHTH degree polynomial. If you're curious, that polynomial would look like this. Do the simplification yourself.

$\alpha ^2\cdot B\ -\ \beta ^2\cdot A\ =\ 0$

$where$
$A = 1-x^2$
$B = 2x-x^2$
$\alpha = x^3-2Ax$
$\beta = (x-1)^3-2B(x-1)$

And I apologize for the differentiation mistake in the last post (and for underestimating the degree of the polynomial--you were right, it had to be an even degree. Turns out it was 8). I hope I didn't waste too much of your time.

I have graphed this. The polynomial has five real zero's, so you'd have to check the value of each one. You must check ALL of them because these are all "critical points" for the lagrange method.

(x^3-2(1-x^2)x)^2*(2x-x^2)-((x-1)^3-2(2x-x^2)(x-1))^2*(1-x^2)

Now, if you graph this, it acts like an odd-degree polynomial. Why? Because the eigth-degree terms cancel out. So this is actually a seveth-degree. Please note that this equation is for one thing only: It tells you the critical points of the Lagrangian. That means, the 5 zeros of this polynomial are the ONLY places where the graph could achieve its maxes and mins. This, I believe, is as close to an answer as you'll get. No exact values here. Go ahead and use Newton's method or whatever.

Sorry for the earlier mistake. Hope this rectifies it.

--Jake

14. Originally Posted by Mazerakham
As excited as I was, it turns out I only gave you guys a minimum. It also turns out that I calculated the gradient wrong, which is embarrassing.
--Jake
So? Only way I've learned anything in the past is by making mistakes.

15. @Wilmer: Yeah I want to know how that works out too. ._.

@Mazerakham: With my method I also got an 8-degree polynomial, and well...in school we get to solve 3-degree equations at most so I guess I've overdone myself a bit... It was fun trying though.

Thanks again for help both of you ^^