A farmer is building a CIRCULAR fence, but it is partially built touching the side of his house.
How to find the optimal ratio between the chord length and the fence length such that the area covered is maximal?
>>7746888
To clarify "a" is the chord length touching his house so there's no fence there.
I'm supposed to solve it in mathematica but there's gotta be an elegant solution to this.
just fucking build a fence, no math needed. you act like white people where the first to make pyramids.
>>7746928
Reported.
a = 0?
a as small as possible?
>>7746936
Probably not, because the area would be smaller with the same " l " which is the length of the fence used.
>>7746942
So what do we have to find? The highest ratio of Area/l, being able only to modify the l/a ratio?
>>7746949
The biggest Area, being able to only modify the l/a ratio.
>>7746949
Do a or l have a fixed length? This problem doesn't seem to be complete.
>>7746952
It's hard af, and it doesn't matter if 1 of them has a fixed length, you need to find the optimal a/l ratio, such that the area covered is max possible
>>7746952
Only way the question makes sense to me is if the area is fixed and you're trying to find the minimal fence length. That's the classic problem.
>>7746955
You fucking retard read the post again, l only covers 1 part of the circumference, rest is a. If you make a full circle then it's a circle with l circumference, otherwise it has a perimeter of a+l and the whole area is bigger.
>>7746929
Egyptians were africans. you are racist dude.
>>7746888
there is no optimal ratio.
While the length increases linearily, the area increases as the square of the length (more or less). So it is always beneficial to increase the length of the fence because the covered surface will only increase even faster.
Or maybe I misunderstood your question
>>7746962
The area of a circular segment is always lower than the one of a circle with the same radius.
>>7746962
so is the fence length fixed or what?
https://youtu.be/ISWGxmWJYjw
>>7746966
The ratio a/l A/L!!!!
If you double both l and a then the area gets four time as large while the ratio a/l stays the same.
There is no optimal solution.
>>7746975
listen you faggot.
if you want an answer to your homework problem, start by BOTHERING to write a correct problem.
The optimal ratio a/l is 0.
Happy faggot?
The area of the blue part is given by:
A = r^2 * (pi - 1/2 * (alpha - sin(alpha) ))
With r = radius; alpha = angle underlying? the chord a
l = r * (2 * pi - alpha)
a = 2 * r * sin (alpha/2)
Now?
>>7746979
Let's say the farmer has a fixed length fence l.
Now he's looking for optimal placement of the fence such that the area is max possible.
The chord "a" represents a side of his house, there is no fence placed along the "a" side.
Now figure out the ratio a/l such that the area covered in blue on the picture is max possible.
>>7746984
Yeah but how does he express the A in terms of l and a?
>>7746987
>fixed length fence l
>The chord "a" represents a side of his house
>figure out the ratio a/l
Are the sides of his house meant to be variable in length?
>>7746991
I can't find a way to do that
Can I do:
(2pi - alpha)^2 = 4pi^2 - 4pi*alpha + alpha^2
?
>>7746994
Why?
>>7746987
a is still 0. Full circle. a/l is still 0.
Impolite sage.
Man I give up I can't find a way to express the area without the radius or the angle alpha, every time I try I just get a longer expression with even more r and alpha.
>>7747010
No it's not you're just wrong.
>>7746967
/thread
The shit is wrong with you niggers
>>7747017
I wasn't even him.
Impolite sage.
Jesus Christ you people are morons. The chord doesn't need fencing because he's using that part of the house as a fence. The circle would then have a larger radius the more house is used.
Learn how to apply your math skills.
>>7747026
Finally someone gets it. I swear /sci/ has gone to shit in the past year-
>>7747026
C++ confirms this
I've lost 100 IQ thinking about this problem
Alpha = pi
So a = 2r * sin (alpha/2) = 2r * 1
L is equal to half the circumference (l = 2pi * r/2), because alpha = pi
So a/l = 2r/pi * r = 2 / pi = 0.63661977236...
Your ratio is 2 / pi
So you want to solve the problem
>max A
>subject to
>A + r^2 * (1/2 * (alpha - sin(alpha)) - pi) = 0
>l + r * (alpha - 2 * pi) = 0
>a - 2 * r * sin (alpha/2) = 0
A, a, r and alpha are variables while l is constant?
Ok first you eliminate A from the system:
>max r^2 * (pi - 1/2 * (alpha - sin(alpha)))
>subject to
>l + r * (alpha - 2 * pi) = 0
>a - 2 * r * sin (alpha/2) = 0
Now eliminate r:
>max l^2 * (pi - 1/2 * (alpha - sin(alpha)))/(alpha - 2 * pi)^2
>subject to
>a + 2 * l * sin (alpha/2) / (alpha - 2 * pi) = 0
You can always choose a to fit the constraint so you can simply remove that contraint:
>max l^2 * (pi - 1/2 * (alpha - sin(alpha)))/(alpha - 2 * pi)^2
Define f(alpha) = l^2 * (pi - 1/2 * (alpha - sin(alpha)))/(alpha - 2 * pi)^2 and solve
f'(alpha) = 0.
After knowing the right value of alpha you can simply calculate a, r and A in order.
>>7747052
To say it in a simple way you just need to build a semicircle (so alpha = pi = 180°).
>>7746888
Intuitively I want to say 2/pi
The problem is wrongly put together. OP is a faggot. Sage and report.
Max area you can cover is a full circle when a = 0.
>>7747152
Nope, you're just retarded. The way I see it you'd have be trying not to see it right, ergo you are a retard.
So you mean the least amount of fence used for the most amount of area covered when building it circular?
If that's not what you mean then I don't get it at all.
>>7747245
If the best ratio is l/a when a approaches l then the most reaaonable solution is a semicircle, if the optimal ratio is when a approaches 0 then a circle is best.
>>7746888
>A farmer is building a CIRCULAR fence
No farmer is building a circular fence, ever, said my Russian colleague, why do your idiot teachers feed you such absurdities? First you believe in absurdities and then you commit your atrocities, that's why..
We do rectangular fences because they follow the secret rule.
The secret rule?
Yes, the secret rule. I'll tell you. Imagine you want 100 square metres in 4 partitions and you build it against a wall. This means that east is two and north is three and a hundred and twenty-four.. which means you need exactly sqrt(2400) meters of chickenwire .. a 50m roll.
You did this in your head?
Sure, it's called the secret rule for a reason.
Damnit, I feel undereducated.
>>7746888
the same way you optimise any problem
now do your own fucking homework
You can parametrize the area with the angle that the intersections of the fence and the wall make with the center of the circle, but differentiating with respect to that parameter gives you a transcendental equation.
[math]
r = \frac{l}{2\pi} [1 - \frac{\theta}{2\pi}]^{-1} \\
and A = \pi r^2 [ 1 - \frac{\theta}{2\pi} ] + ar\cos{\theta /2}
[/math]
It's not as easy as you think.
lets try this one more time.
[math] r = \frac{l}{2\pi} [1 - \frac{\theta}{2\pi}]^{-1} \\
A = \pi r^2 [ 1 - \frac{\theta}{2\pi} ] + ar\cos{\theta /2} [/math]
>>7748306
>>7748307
didn't render at first for me but I guess it was okay.
As pic related shows, something of this form should have a maximum at theta less than pi. Not sure exactly, but I wouldn't be surprised if it's close to pi/2.
Then knowing that solid angle and basic geometry, it should be easy to find the ratio you want.
>>7746969
Yes, but the amount of wall is not.
>>7746952
I'm interpreting it as length stays constant.
does nobody on this board know Calculus of Variations?
I mean, I forgot how to do it entirely, but that's how to do what OP is asking, it's a minimization problem with a boundary condition
>>7750097
I suggest you actually try the problem, it's not analytically solvable.
>>7750127
I think it's just a stupid trick question.
The ideal shape for maximum area from a fixed path length is a circle.
Therefore, the best ratio of circle to chord is as little chord as possible.
Let l = 1
Let p be the fraction of the circle's diameter covered by l
1 = p 2pi r
r = 1/(p 2pi)
σ = p 2pi
A = (2 pi p - sin(2 pi p))/(8 pi^2 p^2)
A' = (cos(pi p) (sin(pi p)-pi p cos(pi p)))/(2 pi^2 p^3)
cos(pi p) (sin(pi p)-pi p cos(pi p)) = 0
cos(pi p) = 0
p = 1/2
So the area is maximal when a semicircle is made, which means
r = 1/pi
σ = pi
a = 2/pi sin(pi/2)
a = 2/pi
So the optimal between chord length and fence length is 2/pi
>>7746992
Yes. This is a bloody highschool calc optimization problem.
>>7747010
Using only a half circle is just as area efficient.
>>7750130
No, that doesn't follow because the wall isn't fenced. You get that bit for free.
