Maximize the area of one ellipse!

[@MichaelPennMath]

Hi all: I made a mistake around here: 18:57 -- classic Michael Penn! It should be straightforward to finish it off correctly. Reply to this comment with the correction!

[@goodplacetostop2973]

It wasn’t a mistake, it was just a happy little accident.

[@romajimamulo]

You forgot the fact your b squared on the right is divided by A squared

[@IAmTheFuhrminator]

I was losing my mind when I noticed this, I'm glad you pointed out this "little accident". Great video overall though!

[@tonyhaddad1394]

It must be 3b^2 not 3a^2b^2

[@Walczyk]

I get a^2*b^2 = 1/12, very cool

[@hocky-ham324-zg8zc]

I hate when people ask me “what is the point of doing math?” My response is always: “y_0?”

[@hocky-ham324-zg8zc]

@3ew you’ll understand the joke or the answer to “what is the point of doing math?”

[@eugenekim3012]

I think I spotted a mistake at 19:30. If we multiply the entire equation by 3a^2, then the final term should be -3b^2 and not -3a^2b^2. With that said, I get a^2 = 1/2, b^2 = 1/6, and my final answer for the area is pi/2/sqrt(3). Another error in your answer is that the area of the three ellipses is larger than the big circle, which is obviously impossible.

[@GiacomoAakbr]

yess, i too caught that error! even the best make mistakes!

[@JB-ym4up]

3 objects of area pi(root(3/4)) do not fit in a circle with R=1.

[@studset]

Here is a picture with your values. imgur.com/a/amWxhKL

[@eugenekim3012]

@@JB-ym4up Actually if you recheck you calculation the total area of the three ellipses is pi*sqrt(3)/2, that's less than pi.

[@JB-ym4up]

@@eugenekim3012 yes, but he gives that as the final answer, the question is the area of one.

[@Milan_Openfeint]

Nice, first time I see an optimization problem where the best ellipse is NOT a circle.

[@thatoneginger]

Man, you really punished yourself at 18:35. You do awesome work. You can just say “or rather,” or something. Trust me, we’re all sitting over here WAY too impressed with how hard you slam these problems to worry about a slip up like that.

[@InDstructR]

Me: Why did you use this method to solve the problem? Penn: y nought?

[@t39an8r]

I chuckled when I wish that I hadn't

[@Carcharoth313]

KZbin: Michael Penn just uploaded a new video. Wanna watch it? Me: Yeah, y_0

[@lien3729]

y_τ

[@madhukushwaha4578]

If you want more harder questions then I will highly recommend you this channel latest videos #mathsandphysicsfun .

[@markkennedy9767]

I like how you motivate and give the why for each step (which a lot of maths expositor's don't do). Makes it so much easier to follow. Good stuff.

[@gastoncastillo9946]

Your videos are quality in its maximum expression

[@goodplacetostop2973]

21:19 ight imma head out 24:11 Hi everyone, I hope you’re having a great morning/day/evening/night. Homework time : Consider two triangles ABC and DEF such that: 1. The lengths of the sides of the triangle ABC are positive consecutive integers and the same property holds for the sides of the triangle DEF. 2. The triangle ABC has an angle that is twice the measure of one of its other angles and the same property holds for the triangle DEF. Compare the areas of the triangles ABC and DEF.

[@srijanbhowmick9570]

I think only one such triangle exists with sides 4,5 & 6 and has an area of (15*sqrt(7))/4 . There will be 3 cases in total which are :---- (i) The greatest angle is twice the smallest angle which leads to the 4-5-6 triangle . (ii) The greatest angle is twice the 2nd largest angle which doesn't lead to a triangle with integral side lengths. (iii) The 2nd largest angle is twice the smallest angle which also doesn't lead to a triangle with integral side lengths. Correct me if I'm wrong!

[@goodplacetostop2973]

@@srijanbhowmick9570 short answer= Actually ABC and DEF are the same triangle. AC = 5, AB = 6, BC = 4, and

[@srijanbhowmick9570]

@@goodplacetostop2973 Yeah, but that was the whole point of my reply! Anyways good question for fooling others!

[@marcpietro2732]

22:24 (even if the expression is incorrect) you don't need calculus to determine the maximum value of b^2 * (1-b^2) because the greatest value of a product whose sum is constant is obtained when both terms are equal, which leads to b^2 = 1-b^2, so that b^2 = 1/2 and b = 1/sqrt 2.

[@ZeonLP]

18:40 existential crisis.

[@JR-zc5pz]

@@dgarrard100 brilliant

[@vikiv.1352]

Thank you! Now I can put 3 *perfectly* thick sausages on my plate. My quarantine became a lot better!

[@megauser8512]

LOL!

[@axiomeVpostulat]

Using tangency between the circle and the ellipses work fine too. Of course we get the same value A=π/(2√3) , if we are not mistaken.

[@akhotaba7866]

Can you make a video solving this difficult diophantine equation? x^y + y^z = z^x Solve for the integers.

[@orangetux]

I really like this problem. Thanks!

[@takyc7883]

Everyone’s pointing out the mistake but why no one points out those biceps :(

[@mathfincoding]

And the back muscles 🥵

[@pokoknyaakuimut001]

So complicated problem, but that's so wonderful ✨😍

[@giacomolanza1726]

The drawing is misleading. If the vertices of the ellipse lay on the circumference, then the three ellipses and the circumference would be all secant. Conversely, if the ellipses and circumference are requested to be all tangent, then the vertices are interior to the circumference.

[@darkshoxx]

19:25 you multiply b^2/a^2 with 3a^2, so the final term should be 3b^2, not 3a^2b^2, right? In Particular you'd have to substiture a^2=u or something similar to get to the correct roots, right?

[@ahmadkalaoun3473]

I think you're right..look how it affects the answer The maximum area of one ellipse is [ √3/4 π] Then the area of the three ellipses is [ 3√3/4 π] Wich is bigger than π wich is the area of the entire circle! For sure that doesn't mean that the video isn't great...

[@dufflepod]

Yep

[@zeitviator629]

Excuse me what does degrees of freedom mean ?

[@darkshoxx]

@@zeitviator629 not sure if you wanted to ask this here instead of the main comments, but the short answer is this: DF=(Number of Variables) - (Number of Equations) and gives you an indication of the dimensionality of the object you're looking at. Take a circle of radius 1 for example, locally it looks like a line, so 1-dimensional. It can be defined as x^2+y^2 = 1, so one equation, 2 variables: 2-1=1 degree of freedom, so the object/solution set is expected to be one-dimensional. If you instead allowed the Radius R to be variable, you'd have x^2+y^2=R^2, and therefore 1 equation with 3 variables, so 2 degrees of freedom, as you can fill the entire plane with circles of every possible radius. 2 Degrees of freedom means that you can "freely" move along 2 independent "axes" in the solution set, in this case, the plane. In the Circle with radius 1, you have 1 degree of freedom, and hence can only move along one (slightly warped) axis, which in this case means "along the circle".

[@zeitviator629]

@@darkshoxx Thank you, sir. I completely get it. I did mean to ask you personally. Thanks again.

[@paolomilanicomparetti3702]

thanks for the fun problem and clear explanation, but i would have liked if you started with some hints to try to work it out myself

[@yaroslavdon]

Like some before me said, there is a problem around kzbin.info/www/bejne/q37XeX1sbLSIrq8 . A sanity check would be 3A π , which means there was a mistake in the calculation.

[@johnloony68]

Your answer = 3 ellipses have an area of 4.08. Correct answer = 3 ellipses have an area of 2.72. Big circle has area of 3.14.

[@danielbranscombe6662]

why did you choose that variable for the center of the circle? Why not? LOL

[@ks000-u9m]

XD

[@asklar]

to find the x of the intersection of the circle with the ellipse, you could just set y=0 since the tangent point will be at y=0 which gives x_t^2+y_0^2=1, and x_t^2/a^2=1

[@marce3893]

the tangent point is a bit below y=0. how much it depends on how fat the ellipse is

[@thatoneginger]

This whole video I was like, “Sure, why not?”

[@tomatrix7525]

Another nice problem for me after school. V. Good

[@ZahlenRMD]

A true master.. More powers master

[@eytanmann6208]

So, as noted there is a mistake - the correct answer is that a=1/sqrt(2) and b = 1/sqrt(6). as such the result of the elipse area is A=pi/sqrt(12). that means that 3 such elpse fills about 86.6% of the circle area - while the result suggested here is not possible as it will not fit 3 elipses into the circle

[@natepolidoro4565]

I got that in general, for a radius r of the circle, the max area of one ellipse is (1/6)sqrt(3)*pi*r^2 .

[@natepolidoro4565]

This gives that the maximum proportion of the circle that 3 ellipses in this fashion can fill up is sqrt(3)/2 -- wow, what a common trig value, which is approx. = 86.60% of the circle.

[@MrRyanroberson1]

beyond the mistake from earlier i would further apply the second derivative test to make sure that the function's slope is decreasing at that point, therefore making that a proper maximum: 2-12b^2 = 2-6 = -4, indeed negative

[@pierre-henrilebrun36]

The A coefficient in the quadratic form is negative then the extremum of the quadratic function is a maximum.

[@tahirimathscienceonlinetea4273]

Isn't a problem bro .I know this is the math ,so a little mistake in calculation is missing the whole thing but sometimes an error isn't an issue because the most important thing the way you solve it which is logical and mathematically acceptable.Anyway,you have been doing a great job keep on

[@damascus21]

Analytic geometry is my absolute favorite math bro

[@demenion3521]

when you solve for a² correctly, you should get a maximum value of pi/(4*sqrt(6)) where b=1/sqrt(6) and a=1/4 (if i didn't make a mistake myself)

[@Milan_Openfeint]

You did ;-)

[@demenion3521]

@@Milan_Openfeint i thought so. somewhere in between my piece of paper was full and i got confused xD

[@ivarangquist9184]

I think this problem is poorly stated. You didn't mention that the equilateral triangle had to be at the center of the circle. Also, what does "each line segment is defining an ellipse" mean? I thought it meant the triangle was tangent to the ellipses, but it seems like you meant the corners of the triangle lies on the ellipses.

[@Denis_Ovchinnikov]

Curiously, if you are solving similar problem for n ellipses, you'll find that always a=b*sqrt(3). Btw, the answer is Smax=2*pi * sin^2(pi/n) / 3sqrt(3). It works even for n=2!

[@famancin]

Unfortunately that's not true if n>4, when the area becomes a strictly decreasing function of a^2: in these cases, the area achieves its maximum, which is pi/(2+m^2)^(3/2), at a^2=b=1/(2+m^2), where m=cot(pi/n). Actually, before maximizing the area, Michael should have checked that the x-coord of the points of tangency between the circle and the ellipse were real. Had he done that he would have found that a^2 must be greater than or equal to 1/(2+m^2). Furthermore, b^2 must be positive and this establishes also an upper bound for a^2: a^2

[@zdrastvutye]

19:55 rifle=flier=refil that's an idea for a combinatoric problem like the german word gartezaun=augenarzt. the trick is "gartezaun" is not spelled correctly but gartenzaun(german=gardenfence (english) and augenarzt=visual doctor or ophthalmologist or betrug=geburt or sportarrten=transporter. these are so called "anagrams" see www.quizknacker.de/buchstabenkombination/gartezaun.html it just gave me "schwalbe" for "blechwas" while blech = tin

[@kavitabani6340]

A similar result can be come upon by joining the centre with intersection of ellipse and circle and then applying pythagoras theorem.

[@Qermaq]

y nought cracked me up

[@CM63_France]

Hi, For fun: 1 "what we want to do is", 1 "the first thing that I want to do is", 1 "so next, what we are going to do is", 3 "great", 1 "so let's may be go ahead and write that down", 1 "may be I'll go ahead and", 1 "so I'll may be go ahead and".

[@fredfrancium]

I like the final code, that is a good place to stop

[@kqp1998gyy]

Great!

[@thayanithirk1784]

It's nice that you revealed the book do this more in your upcoming vedios it would be nice and solve problems from Titu andresccu book for number theory

[@MultiNeurons]

Yes: it should be: a^4 = 3*a^2 - 3*a^4 - 3*b^2 (instead of 3*a^2*b^2)

[@eytanmann6208]

highly likely a mistake in the result. The area of the circle is simply pi as R=1. if we take the final result of the elipse to be A=pi*sqrt(3)/4 and we need to fit 3 such into the circle the area of the 3 elipses will be LARGER then the circle - obviously can not be. as the area of 3 such must be SMALLER then the circle... I think I spoted the mistake it is at 19:20 time of the movie --> the eq (a^2)/3 = 1-a^2-(b^2)/(a^2) is multiplied with 3a^2 will result with (a^4) = 3a^2-3a^4-3b^2 that can be further simplified but easier to extract b as function of a (and not a as function of b ).... obviously results with totally different answer

[@koenth2359]

It was not the best place to stop. A simple sanity check would have revealed that the 3 ellipses (as calculated) have a combined area larger than the circle.

[@CTJ2619]

i would like to see the original Japanese text of this problem

[@physicsjeff]

Like the ancient Sumerian proverb says: 3 identical ellipses instead of one big circle gets the job done 86.6% of the time.

[@jackrubin6303]

I get 4(a^4) - 3(a^2) + 3(b^2) = 0. So df(a)/da = 16(a^3) - 6a = 0. Hence a = (6^0.5)/4 and b = (3^0.5)/4 and Maximum Area of each Ellipse = 3pi/(8*(2^0.5)). Does anyone agree? Michael Penn what do you think? Jack

[@jackrubin6303]

Then the maximum area for all 3 ellipses together is 2.499 square units which is less than the area of the circle which 3.14159. Jack Rubin

[@craig4320]

Wondering whether Lagrange Multipliers would work?

[@xCorvus7x]

10:29 Is it a coincidence that y0 has the same value as the coefficient A from the elliptical equation where you had multiplied out the part from the linear equation?

[@sirlight-ljij]

I don't really see why is the problem stated as "maximum" ellipse area, what is the parameter we are maximising over? Isn't there a unique arrangement of inscribed ellipses like that?

[@ScottNesin]

There are infinite ellipses that can be inscribed that way. You can inscribe three equal circles (ellipses where b=a): qph.fs.quoracdn.net/main-qimg-ff688f0ed0713d3a3ef7b4c857c9d9de.webp You can inscribe three equal lines (ellipses where b=0): etc.usf.edu/clipart/43400/43436/3c_43436_lg.gif And you can inscribe everything in between. Think of it as a function b=Q*a, where Q=1 in the circle example above, and Q=0 in the line example above. There is some value of Q between 0 and 1 inclusive that yields an ellipse with the greatest area. At 20:03 he expresses the relationship between a and b in a function that would define what Q gives the maximum area (but notes in the pinned comment that it is incorrect). It is beyond my ability to simplify that function, but if you read the comment by Terry Ligocki , he says the maximum area will be when Q=1/sqrt(3) .

[@MrWarlls]

The circle is almost perfect. But, there is still work for the ellipses. 😉

[@MichaelPennMath]

I use a string to draw the circle, but I don't have a technique for drawing an ellipse. Any suggestions??

[@gastoncastillo9946]

@@MichaelPennMath maybe you could fix the foci in the ellipse, and use a string attached to those foci and as the sum of the distances from a point to the foci is constant then you can change it by just changing the size of the string

[@MrWarlls]

@@MichaelPennMath , a string fixed to two differents points.

[@tonyhaddad1394]

Copy a picture of an ellipse and put it on a wooden plate and cut this wooden to get a ruler of ellipse 😅

[@Fun_maths]

@@MichaelPennMath if you're fine with two pointy objects on your board then attach them and they will be the focus points and take a string passing through both objects, stretch out as far as you can and draw

[@byronwatkins2565]

b^2/a^2 times 3a^2 is 3b^2 not 3a^2 b^2. Max area = pi / 2 sqrt(3)

[@michaelgian2649]

20:16 The "major" and "minor" axes referred to can more conventionally be called "semi-major" and "semi-minor"?

[@michaelgian2649]

Shades of Matt Parker

[@SwordQuake2]

19:28 that's not true. Should be 3b^2.

[@einzigermylee5996]

Assuming this is a 3-D problem, and the ellipses are circles forming 3 sides of a cube (one corner is in the middle and the 1-circle is a sphere cutting the inner circles of the sides of the cube, which are the ellipses), this can be solved using only pythagoras. But i guess this is too boring. ;)

[@leif1075]

How do tiu knownthe line from the corner of the triangle to the center of the circle bisects the angle?? You dont know that for sure at 4:21??

[@paolomilanicomparetti3702]

due to symmetry: there are three equal ellipses in the circle, the figure is symmetric around that tangent

[@recklessroges]

How should I know where to put the origin? (I would have set the origin to the centre of the circle, as my first guess.)

[@ipudisciple]

Me too. There’s no universal answer to this question, though. Both work, and intuition hopefully guides which will be faster.

[@meiwinspoi5080]

taken down? hope u re-upload.

[@dheerdaksh]

23:40 _Nice_

[@drchaffee]

Japanese temple geometry relied on calculus?

[@sergiokorochinsky49]

So many thing to choose!... Why not square... why not x... why not minus a... why not this... why not that... Ohhh... wait... maybe... ...is he talking about y_0?

[@InverseTachyonPulse]

Perfect circle, not so much the ellipses 😅

[@stvp68]

I don’t understand how the triangle’s sides define the ellipses if they aren’t the axes.

[@studset]

@Michael Penn - when the osculating circle is bigger than the circumscribed circle then there is six points of contact but when it is smaller there is only three. Illustration in the linked image: imgur.com/a/2gUncLM

[@DS-xh9fd]

At 22:50 you ignore the fact that 0 is also a solution to the equation. Of course, 0 gives a minimum, not a maximum, but you still should check.

[@hadiedan8515]

Area= pi/12^0.5

[@vladimirLen]

The title is misleading. It implies that the maximal area of one ellipse is not the shape of a circle.

[@AA-gw6wd]

I think you will convey a more professional and confident attitude if you don’t stop and get frustrated with yourself every time you miss speak, no ones going to hold it against you anyway.

[@xevira]

"I'm going to call the center of this circle 0 comma y naught." "I dunno, '0 comma' seems like a weird name for the center of a circle."

[@Nik12251114]

Being a medical doctor this just reminds me of the cross section of a penis. Good to know the max area now.

[@ethancheung1676]

Those are sausages

[@vaxjoaberg9452]

In mathematics we call them sausagoids.

[@taransingh5026]

hey micheal PLSSSSSSS can U just solve 2020 JEE ADVANCED maths paper................I REALLY LIKE YOUR TEACHING PLSSSSSSSSS..

[@hamiltonianpathondodecahed5236]

wait a min , I am early and this one is fun