More Minimising Without Calculus

We minimise a^2 + b^2 + c^2 + d^2, subject to the constraint abcd = 9.
00:00 Intro
00:20 2 variable problem
03:56 Building on this case
05:00 Considering negatives
06:00 3 variable problem
10:03 4 variable problem

[@RanOutOfIdeas4aName]

AM-GM inequality! Take the numbers a^2, b^2, c^2, d^2. They are all non-negative, a^2, b^2, c^2, d^2 >= 0, so we can apply the AM-GM inequality. AM >= GM -> (a^2 + b^2 + c^2 + d^2)/4 >= fourth_root(a^2 * b*2 * c^2 * d^2) = fourth_root( (abcd)^2 ) But we are given abcd = 9, so fourth_root( (abcd)^2 ) = fourth_root(81) = 3 Plugging this in and multiplying both sides by 4, we get: a^2 + b^2 + c^2 + d^2 >= 12 The minimal value happens at equality, and with a = b = c = d, so the answer is 12. Indeed, abcd = a^4 = 9 -> a^2 = 3 -> 4*a^2 = a^2 + b^2 + c^2 + d^2 = 12. Your content is awesome! I’ll watch and see how you did it later, Dr. Baker.

[@numbers93]

I think that once you have established the 2-variable case, you can generalize to an n-variable case. You can show that for any pair of numbers that are not equal, you can get a smaller square sum by setting them equal to each other.

[@RGP_Maths]

0:46 I would instinctively tackle the 2-variable case by saying a²+b² = (a-b)²+2ab = (a-b)²+18, which is minimised when a-b = 0. BUT I appreciate that you used the longer approach to introduce the technique you were going to need for 3 or 4 variables. However, for people who have learned the AM-GM inequality, the post by @RanOutOfIdeas4aName really is the proper way to solve the whole problem!

[@user-cd9dd1mx4n]

Organised and systematic! I really like your wonderful videos. Please, never stop.

[@magicmeatball4013]

The best time of the week once again

[@yoavshati]

You can visualize this as minimizing the length of the greatest diagonal (squared, but that doesn't matter for finding the minimum) in a cuboid with a fixed volume

[@MrHenryG123]

Intuitively got 12 instantly. The symmetry of the equations (i.e. order does not matter for either equation), minimizing can be done by just picking the minimum value of each, which occurs when setting all values equal to each other (i.e. a=b=c=d=sqrt(3)), which gives 12 for the answer.

[@MrSnic]

Seems weird that you decided to use the two-variable argument for the two first variables, but it seems to be easily expanded to the whole a^2 + b^2 + c^2 + d^2, by saying that having any two different variables would result in having another better solution

[@DrBarker]

This is a good point, and an interesting direction to take the proof! We can make this rigorous by saying that whatever our starting values of a,b,c,d are, we can first replace them by sqrt{ab}, sqrt{ab}, sqrt{cd}, sqrt{cd}, then make the other pairs sqrt{ab} & sqrt{cd} equal, to get sqrt{abcd}, sqrt{abcd}, sqrt{abcd}, sqrt{abcd}. If we want to generalise this to go from the 2 variable case to n variables, perhaps we could even use properties of the geometric mean like GM(a,b,c,d) = GM(GM(a,b),GM(c,d)) to help.

[@MichaelRothwell1]

Exactly what I thought, with the added benefit of this method extending to n variables. It's intuitively obvious that the result must extend to n variables, but I have the feeling that the method in the video is likely to run out of steam after n=4, due to the probable difficulty in analysing the polynomial you get after taking out a factor of (λ-1)².

[@nicholasmoffett4327]

Yeah, I thought that was the entire reason for the two-variable example... the rigorous explanation of the two-variable case trivializes any n-variable case by showing that inequality indicates non-minimality, which made me not sure if I was following when he started factoring everything!

[@jacemandt]

In the 2-variable case, you can do a simple variable substitution of a=3a' and b=3b' to simplify the problem to a'·b'=1 instead of ab=9, because this doesn't change minimization question. But the graph of xy=1 is clearly closest to the origin at (1,1), which is what x^2+y^2 is really measuring. I imagine these hyperbola-type graphs of "product=1" behave the same way in higher dimensions. If you can prove that, you can turn this into a graphical/geometric problem instead of an algebraic one.

[@botot0]

QM-GM inequality solves it in two steps, quadratic mean is always greater than or equal to AM which is greater than equak to GM, sqrt((a^2+b^2+c^2+d^2)/4) >= fourth root of abcd. Square both sides x/4 >= sqrt9 x>=12

[@jamesstrand123]

I more so simplified the problem as a=b=c=d, then solve a^4=9, a = 4th root of 9, same as sqrt(3), I had found similar problems like this one, and just point to the fact that all variables can be equal to each other

[@tunneloflight]

Given the problem as stated, i.e. not constrained to integers or reals, the minimum occurs where a -> 0i, while b, c, and -> infinite-i and where |a| = 9/ |bcd| = 9/ |d^3|. The minimum sum then approaches a transfinite negative value.

[@Bayerwaldler]

First guess (b/o symmetry): a=b=c=d=sqrt(3). Sum of the squares = 12. Can‘t be any other way - can it?

[@OldTiredFat]

No need to prove the three minimum. You can prove the third by Pythagoras.

[@jursamaj]

My take was that a^2+b^2+c^2+d^2=r^2. r^2 is monotonic in r, so they have the same minimum. abcd forms a hyperboloid of 8 sheets, which comes closest to the origin (minimal r) when a=b=c=d, up to signs.

[@spitsmuis4772]

Are there similarly symmetric problems that are not absolutely minimal at equality? (I guess the symmetry will always guarantee a local minimum/maximum?)

[@callumyoung7785]

maximise abs(abc-0.5) subject to a^2+b^2+c^2=3 and a+b+c>0 has neither absolute min or max at equality. max when one of them is minus 1, others are one. minimum asymetric too

[@SierraHunter2106]

We can us the am > gm inequality

[@Hipeter1987]

In the last section, shouldn't it be lambda^2, not lambda?

[@MagicProG]

Why didn't just use Lagrange method?

[@theCDGeffect]

why not look at all 4th roots of 9? i*sqrt(3) also works here, and would be the true min unless there's a reason to omit complex solutions.

[@DrBarker]

This is an interesting follow-up question. If we allow complex numbers, we could take 3, -3, i/x, and i*x for any real number x. Then the sum of squares is 18 - 1/x^2 - x^2, which we can make arbitrarily small by picking a really big value of x. So if we allow complex numbers, there would be no minimum!

[@theCDGeffect]

​@@DrBarker interesting, that's actually pretty cool, thanks for the follow up

[@Axl12412]

Ecclesiastes 7:3 Sorrow is better than laughter; it may sadden your face, but it sharpens your understanding. ‭Proverbs 14:13 Laughter might hide your sadness. But when the laughter is gone, the sadness remains. ‭Proverbs 25:20 Singing happy songs to a sad person is as foolish as taking a coat off on a cold day or mixing soda and vinegar. ‭Ecclesiastes 5:5 It is better to say nothing than to make a promise and not keep it. ‭1 Corinthians 15:34 Come back to your senses as you ought, and stop sinning; for there are some who are ignorant of God-I say this to your shame. ‭Proverbs 16:24 Kind words are like honey-sweet to the taste and good for your health. If you want to know the Truth follow Jesus.

[@de_oScar]

Euler > > > Jesus

[@JoseAntonio-ng5yu]

It'd be easier with Calculus. BTW, guessing from the start that the solution should be the variables being equal and working around the idea is questionable

[@erfanmohagheghian707]

All this work to avoid Lagrange multipliers? Lol