Cambridge Interview Question: Trust your Intuition

[@joseph-fernando-piano]

The way I would visualize the problem is by looking at the quantity (ab + bc + cd) as the "area of a rectangle with sides (a + c) and (b + d), minus the area of the rectangle (ad)"; to maximize the expression of the area, we should minimize the size of ad, hence a = d = 1 since all four variables must be natural numbers. The problem then becomes: a rectangle has sides x and y (x = a + c, y = b + d) which sum to 63 and are natural numbers, maximize the area of the rectangle, which is done for a square shape. Since the perimeter sum is odd, the closest we can get is a rectangle with sides 31 and 32, therefore b = 30, c = 31 (or vice versa), and the maximum of the product is (1*30 + 30*31 + 31*1) = 991

[@adityarajsingh3299]

Wow it really is a awesome way to visualize and very easy to understand

[@h0tp]

yes amazing way to think about it thanks for sharing

[@aipause]

Dat solidwork dude

[@SaiTeja-ls3ns]

🔥

[@remy5749]

Ok, but for me the max is 992,25 because you are 4 values a,b,c,d and there are égals to 63 so if you want the max for ab+bc+cd you want to find the max value to ab or bc or cd and 63/4 = 15,75. So 15,75*15,75 = 248,0625 this a max you can have with the « limit a+b+c+d=63 » so when you use this 248,0625*4 is 992.25 so the max is this not 991

[@cakeboy3351]

Basically the comment section is divided into two groups : those who count 0 as a natural number and those who don’t.

[@adampaul7905]

Yes it's insain ...

[@joe_z]

Since this is an interview, I would have taken the time up front to ask whether or not they include 0 or not.

[@thelittlegumnut]

So basically, if Cambridge doesn't like you for some arbitrary reason, they'll just switch the standard they're using for the set N and claim you're wrong on that basis lmao.

[@rbanerjee605]

Nope, if Cambridge don’t like you for some arbitrary reason, they will not invite you for interview.

[@vimanmandala5444]

@@rbanerjee605 savage

[@jpa_fasty3997]

Point isn't to get the answer correct necessarily in the interview, it's to show your approach to solving it. That's what they're interested in.

[@iainprendergast8311]

In your arrogant opinion

[@donaldli4755]

@@jpa_fasty3997as if 😅

[@eug_499]

A shorter approach is based on the symmetry of parabola. The product of two real numbers with fixed sum is at max when the numbers are equal. For 63, they are 31.5. The closest symmetric natural numbers are 31 and 32. Product of smaller numbers declines much faster than the two variables. Based on this and on the symmetry of parabola, the numbers are 1, 31, 30, 1. The result in question is 991.

[@ShiroiTensh1]

Btw, I'm aware that some countries exclude 0 as natural numer, but, a=d=0 and b=32 c=31 gives max of 992

[@aromeran]

Values excluding 0 are: (1, 30, 31, 1), (1, 31, 30, 1) -> 991 Values including 0, there are 128 groups

[@faheem4581]

@@walterbolduc4744 bruh

[@faheem4581]

it adds to 63

[@gabrielf66]

if it was excluded, it would be a, b, c, d E N*

[@walterbolduc4744]

@@faheem4581 there was another comment where it added up to 64 but I guess it got deleted

[@Deus_Auto]

This is like Mind Your Decisions except that it gets to the point without much padding for time.

[@mcwulf25]

I used Lagrange multipliers, minimise with one constraint. We quickly find b+k=0, a+c+k=0, b+d+k=0, c+k=0. Adding the middle two together and substituting the original eqn gives 2k+63=0 so k= -31.5. So we have b=c=31.5 and a=d=0. Obviously we now have to tweak this for natural numbers which gives the same answer as the video. It's possible to use inequality constraints but that's over-engineering the solution.

[@simonjones9254]

what's your academic background that gave you intuition to do that?

[@jasonwong2242]

@@simonjones9254any multivariable calculus course will teach Lagrange multiplier, and there is also a constraint and a function to optimize in this case.

[@almscurium]

@@simonjones9254calculus 3

[@Analys1s]

I actually thought of doing the same thing. What was your constraint?

[@Handballplayers1]

@@Analys1s I think u would set it up as ab+bc+cd-L(a+b+c+d-63), Solve A: b=L Solve B: a+c=L Solve C: b+d=L Solve D: c=L Then inset the values in the contraint 63=a+b+c+d, 63=L+L, L=31,5, then B=31,5, C=31,5 and then A and D must be 0 This would give 31,5*0 + 31,5*31,5 + 31,5*0 = 992,25

[@metinersinarcan92]

The first thing that comes to my mind is using "perturbations". Assume that a, b, c, d maximizes our objective function ab+bc+cd under the constraint a+b+c+d=63. Our main argument will be that if we change a, b, c, d, the value of the objective function shouldn't increase because otherwise, our choice wouldn't be maximal. Now, if we increase c by 1 and decrease a by 1, the new value of our objective function becomes (a - 1)b + b(c+1) + (c+1)d = ab + bc + cd + d but as we said our objective function shouldn't increase so this implies d 0 then d=0. Similarly, if we increase b by 1 and decrease d by 1, the new value of our objective function becomes a(b+1) + (b+1)c + c(d-1) = ab + bc + cd + a. So by the same argument, we proved the statement: if d>0 then a=0. Our two proved statement together implies that we have a=0 or d=0. Notice the problem has a certain symmetry: If we interchange a with d, and b with c both the constraint and the objective function remains the same so without loss of generality, we can assume that d=0. In summary, we have shown that there exists a choice a, b, c, d with d=0 solving our maximization problem. Then substituting d=0, our problem reduces to: maximizing (a+c)b under the constraint (a+c) + b = 63 or renaming n=a+c for simplicity: maximizing nb under the constraint n+b=63. We immediately recognize this problem and know that the solution is (n, b) = (31, 32) or (32, 31). Then the maximum value that can be obtained by our objective function is 31*32=992.

[@karigucio]

such a solution: in positive real numbers the equation is symmetric, substitute b=c, a=d. Get max(...)=max( 2ab+b^2 ) = max( (a+b)^2 - a^2 ). Since a+b=63/2, we get max by making "a" the smallest (that is 0, not the 1 nonsense from the video ;]). Now we wave hands about how the solution in naturals needs to be around that solution (right, can that work some way?)

[@farhatali2221]

I liked the intuitive approach .. Thanks a lot

[@jongyon7192p]

By KKT conditions of optimality, the gradient conditions is grad f(x) + L grad g(x) = 0, where L=

[@jongyon7192p]

@@raffaelevalente7811 yo i DID get 992 m8! LOL

[@raffaelevalente7811]

@@jongyon7192p I'm sorry, I misread :)

[@TheNothingNihilates]

For those who prefer geometric intuition: think of the rectangle with sides of length a + c and b + d. Perimeter P = 2a + 2b + 2c + 2d = 2 * 63. Area A = ab + bc + cd + ad. Remove the corner that corresponds to the ad term, to get an L shape. Perimeter P' = P = 2 * 63. Area A' = ab + bc + cd. Maximize A'. Whatever their values, taking one from either a or d and adding it to b or c will increase area A' while keeping the perimeter constant. Keep going till a = 1 and d = 1. Whatever their values, taking one from the larger of b or c and adding it to the smaller will increase area A' but keep the perimeter constant. Keep going until they are equal or differ by one. But why 63 though? Feels like it should be 63 = 8^2 - 1, with the 1 being the missing corner with area ad = 1, but that isn't how it works. Aesthetically unpleasing!

[@MichaelRothwell1]

Nice, but still need to show b & c must be as nearly equal as possible. I think you can use the same kind of argument.

[@TheNothingNihilates]

@@MichaelRothwell1 Last 2 sentences, 1st para.

[@MichaelRothwell1]

@@TheNothingNihilates my bad.

[@TheNothingNihilates]

@@MichaelRothwell1 hey, no worries.

[@xyz956]

This is just op.never dreamt of this type of geometrical approach

[@thijsg717]

I wasn't entirely sure whether the 'N' meant 'all numbers' or 'only positive numbers'. If you allow for all integers, the max is infinity. You essentially either use the a or d as a postive, whilst allowing all other numbers to be negative. This way, you can create 'bigger' numbers (read, further away from zero) out of thin air (the a or d compensates for this), thus allowing for bigger positive products that will eventually outweigh the a times b or c times d (depending on whether the a or d is going to be negative) significantly.

[@F_A_F123]

ℕ means natural numbers. It's either nonnegative integers or positive integers

[@thedude5191]

If we consider c=d=0, we have 31*32=992, so what prevents us from doing that? a,b,c,d belong to N so 0 is included

[@vasvas8914]

Nope, zero doesn't belong to natural numbers

[@lachlanwilson1519]

This isnt a Cambridge interview question, its from the Australian AIMO olympiad

[@vlix123]

lmao, I remember taking the AIMO that contained this question. Still, I guess Cambridge could have just recycled it. There’s really no need for interview questions to be original, so long as they are obscure enough.

[@drewkavi6327]

@@vlix123 I doubt this is cam interview question, they tend not to ask questions of this style .

[@geneyoungdho]

How about Lagrange multiplier(Δf = ΣλΔg, just assume a+b+c+d = 63 as g, ab+bc+cd as f?

[@tejasappana4097]

Yup thats what came to my mind for the formal approach

[@jokubaszitkevicius8243]

From the very start it looks as if you need to solve for maximum value of symetrical sum of ((ab)^1*c^0), where a=d. So that is how I would start approach and only now try to prove that a=d.

[@jackflowt]

The maximum is 992 which is the product of 32 and 31 while the rest is zero. You could use geometry to prove this by finding the max area of a rectangle.

[@MilesHemingway]

0 isn't natural

[@jackflowt]

@@MilesHemingway0 is a natural. It belongs to N. You're talking about N*

[@jackflowt]

Check wikipedia for N

[@MilesHemingway]

@@jackflowt I think some professors will define it to include 0, but most I've been involved with have not, as is the case here too

[@jackflowt]

a, b, c, d belong to N={0,1,2,...} I hope youtube does not delete my comments again.

[@coco805]

So two of the terms are used twice, B and C. So it makes sense to maximize those so the BC term is large, and thus minimize A and D. So if A=D=1, B and C are 30 and 31.

[@Sky11631]

But isnt the maximum to just 31*32=992, since an addition will never give more than the "multiplication shift" (like 30*31 = 30*30 + 30*1). So id assume your solution at start is only valid for even integers in equation 1? Or do you use N = N+ not N = N_0?

[@nexusoz5625]

a and d has to be a positive integer, so non zero

[@Sky11631]

@@nexusoz5625 Well is it specified somwhere? If he said it, I missed it. If not it just is not clear, since the natural numbers can mean both: with and without zero. (N_0 or N^+)

[@spitfirerulz]

Lagrange multiplier method where f = ab + bc + cd, and g = a + b + c + d - 63 = 0. Solve for real numbers, and we get lambda = 31.5. Then modify for constraint of natural numbers.

[@pastaliafilamel5831]

Notice that there's another loosely restriction that a,b,c,d>=0, or ab+bc+cd is unbounded (a,b,c,d=-x,-1,1,x+63). It is possible that this restriction is saturated somewhere else if the target function is not this nice.

[@sbares]

That happens to work here, but you would have to prove that the maximum over N^4 is in fact the nearest point to the maximum over R^4. This is the case here, but can fail catastrophically in general.

[@suuujuuus]

Im not a mathematician so my approach is very simple: Imagine we have an allocation for a,b,c,d so that the first equation holds. We can always increase the value of the second inner term, by taking a 1 from a,d and give it to b,c due to the number of occurences + everthing >0. Which gets us to d=a=1 and max(b+bc+c). For bc increases stronger than b+c (b,c>2 => bc>b+c) & max(bc ) min(b-c) for b+c=61 => b=30,c=31 or vice versa.

[@mycrushisachicken]

how is this any simpler than the first aproach in the video?

[@yesdcotchin]

​@@mycrushisachickenit fleshes out the thing which the creator just calls intuition

[@mycrushisachicken]

@@yesdcotchin well, it isnt simpler and im not even sure if what he wrote makes any sense

[@ignaciobenjamingarridoboba2071]

Using a geometry model as a rectangle, i assume a,b,c,d must be different of zero (a,b,c,d must be at least 1). then the answer is 991. Of course, you can assume in fact a,b,c,d could be 0, changing the sides of the rectangle to something more trivial, and the answer would be 992 (31*32). perhaps we need more information to assume if a,b,c,d have a min value.

[@divyanshshukla6059]

Bro ur fan nd subscriber from India 🇮🇳 lots of love!❤❤

[@andrea-mj9ce]

2:30 You should justify the 31 * 32. By consider the function n -> (63 - n)n

[@madghostek3026]

My way was to take arbitrary a,b,c and d, then notice that if the expression is ab+bc+cd, taking 1 from a and adding it to c will be (a-1)b+b(c+1)+(c+1)d = ab+bc+cd-b+b+c= (ab+bc+cd)+c. This means that no matter the actual values of a,b,c,d, doing this operation will increase expression by c, so it's always worth doing it. Same thing applies for pair d and b. So starting from any values you will always end up at a=d=1.

[@BK-qp4uq]

A = D ?! Thats not math. Math would be A=A and never A= anything else ! Another letter means another number. You failed.

[@madghostek3026]

@@BK-qp4uq this is just shorthand way, you know what I mean

[@hongruicui2806]

never realized before that an ipad can be this useful when doing screen sharing in virtual conferences.

[@kenlandon6130]

Is this a lagrange multipliers?

[@ytwestlake]

At the risk of bringing a linguistic argument to a maths fight, surely if a=d then d is redundant and the equation would read 2a+b+c. If d is distinct enough to warrant a different signifier, it has to be distinct from a. The minimum value for a+d must therefore be 3, ad=2...

[@EW-mb1ih]

According to the international standard, ISO 80000-2 which defines quantities and unit used in mathematics, Natural Numbers include 0. In the “Symbols and expressions for standard number sets and intervals”, it clearly indicates that the symbol N means “the set of natural numbers, the set of positive integers and zero” and gives example such as : N = {0, 1, 2, 3…} and N* = {1, 2, 3, …} So my answer would be max(ab + bc + cd) = 992

[@ngc-fo5te]

No.

[@EW-mb1ih]

@@ngc-fo5te Thank you for your thoughtful and constructive response.

[@CromulentEmbiggening]

That's wrong because the ISO 80000-2 uses the wrong definition of natural numbers.

[@EW-mb1ih]

@@CromulentEmbiggening so what source gives the « good » definition of natural numbers ?

[@neilgreen3829]

It's a couple of seconds of intuitive geometric visualisation to figure this out. Also, if you're aiming to maximise ab + bc + cd you can get a larger sum if you include 0 as a natural number, so set a and d to zero unless you're given a definition of natural number that doesn't include zero.

[@terminalrecluse]

Whoa whoa whoa anyone else feel like he skipped a ton of steps at 3:01 ?

[@blueberry23]

Not exactly, for (n-x)x, maxima happens right in the middle. So if n is even, x is n/2 if it’s odd, it’s that way.

[@petrkoucky8565]

What if a and d are 0? Then C and B are 31*32=992 not mentioning a and c could have negative values... then It could be even more

[@thelittlegumnut]

That's what I thought but it turns out a, b, c, and d have to be in the set N which is defined as the set of natural numbers, according to the question. The set of natural numbers excludes 0. Not sure about the negative numbers.

[@thelittlegumnut]

Actually, I see what you're saying about the negative numbers now, but they're still not included in the set N anyways. Good thinking though.

[@Dawlada]

Actually 0 is not an element of Natural numbers, but the explanation is slightly note right, though the answer is right.

[@harounfarhani2138]

@@Dawlada in my country (the Netherlands) 0 is considered a natural number. Something like that should be clarified in such questions to avoid unnecessary disadvantages.

[@Dawlada]

@@harounfarhani2138 i know some mathematicians claim 0 is a natural number but never considered as natural number. In maths we have a lot of misconceptions and misinterpretation but mathematician don't care those unnecessary concerns and conception.

[@TIXU-j5u]

The formal approach seems flawed. In (63-(a+c))(a+c)-ad you cannot assert that the equation will reach max when a is minimum, because a is also factored in the front therefore it is not absolutely clear how a affects the equation.

[@PoliphiloShek]

Should a b c d are different numbers?

[@EnterJustice]

No. If you paid attention, you would have noticed his solution has a = d.

[@CorelUser]

@@EnterJustice if you weren't such a dick, your mum would've loved you

[@issac1071]

Hello! May I ask what book would you recommend to learn this type of mathematics?

[@francoprins8928]

Using multivariable calculus I beleive (a,b,c,d) = (0, 31, 31, 1) is the true answer as it provides ab+bc+cd = 992

[@Robert_H.]

c = 63 - a - b - d max(ab + 63b - ab - b^2 - bd + 63d - ad - bd - d^2) = max(63(b+d) - b^2 - d^2 - 2bd - ad) = max(63(b+d) - (b+d)^2 - ad) = max((b+d)*{63 - (b+d)) - ad) We choose: z = (b+d) < 63 = max(z*(63-z) - ad) We define: f(z) = -z^2 + 63z - ad f'(z) = -2z + 63 f'(zE) = 0 => zE = 63/2 f''(zE) = -2 < 0 => zE is Max z = {31, 32} We make ad as small as possible: a = d = 1. Solution: a = d = 1 b = z - d = {30,31} c = 63 - a - b - d = {30,31} max = 991

[@sunsonny9132]

I used lagrange multiplier first which I get b = 31.5 and c = 31.5, then I looked at the condition which should be all natural number and guessed the answer.

[@B_u_L_i]

Damn I wrote some bullshit on a piece of paper, thinking I'm on the completely wrong path and I'm genuinely surprised I got 991 as well😂

[@pedrojose392]

Good evening! In brazilian concept, zero is a natural number. When talking about strictly natutal we have to symbolize |N* or |N -{0}. I had a teacher that begins his books with chapter zero, to reforce that zero is a natural number. It is interesting So we have for a=d=0 and b=31 and c=32,e.g., (ab+bc+cd)=992>991. Several points of view, it is interesting. I think in France zero is also a positive and a negative integer.

[@MichaelRothwell1]

@@prospektnova9004 "in some other countries" would be more accurate. In the UK 0 is generally considered to be a natural number.

[@ryanjagpal9457]

I didn’t understand anything you said with ab+bc+cd why is it 992>991? What is chapter zero, cause i was thinking it was the front cover of the book

[@pedrojose392]

@@ryanjagpal9457 Good Morning! What I mean is that the answer in the video does not consider 0 as a natural number. And in Brazil, my country, we consider 0 natural. If we can use the zero if a=b=0 and b=31 and c=32 we have that a+b+c+d=63 and (ab+bc+cd)= 992 > 991 the maximun found on the video. And the answer becomes diferent. What I tried to emphasize is that the concept of 0 be or not be a natural number is not pacific. The chapter zero is for the teacher I mentioned what is usually called one.

[@ryanjagpal9457]

@@pedrojose392 Good Morning to Brazil! Yeah I kinda get it now, except the part of cd If ab+bc is 992, then what is d in cd? Why is there a crocodile sign btw isn’t it add?

[@pedrojose392]

@@ryanjagpal9457 , I made a mistake a=b=0 c=31 and d=32. Sorry!

[@JohnSmith-vq8ho]

Lagrange multipliers?

[@tejasappana4097]

Thats what i was thinking

[@delanym]

Pure sorcery how you got the first line of the formal approach?!

[@teckpuiloy]

A geometric approach max area [ab+bc+cd] = max area [ (a+c) * (b+d) ] - min area [ad] [ ad ] [ cd ] | d | + [ ab ] [ bc] | b ------------------- a + c

[@leecherlarry]

Maximize[{a b + b c + c d, a + b + c + d == 63, a > 0, b > 0, c > 0, d > 0}, {a, b, c, d}, Integers]

[@ryanjagpal9457]

That’s a lot of brackets

[@leecherlarry]

@@ryanjagpal9457 parantheses, brackets, braces🎾🍻😋

[@juanjuan5698]

No lie I solved it in about 15 seconds. Im very happy

[@atropineman3541]

Probably should disclose whether any two variables can or can't be equal

[@flyingpenandpaper6119]

Why wouldn't two of the variables be allowed to be equal?

[@lksclaudino]

If it’s not saying a b c d are diferents, we have 3(a^2) = 3 (63/4)^2

[@BucifalulR]

No. They are natural numbers.

[@Gnefitisis]

Perfect. I would have gotten into Cambridge with intuition.

[@samibaheru4029]

Since every notation is different,that is a,b,c,and d couldn’t be the same value.Max would be 963.The other way round 29+31+2+1=63.Now find ab+bc+cd=(29*31)+(31*2)+(2*1)=899+62+2=963.

[@erdes4512]

for a,b,c and d all different integers (not null), max is 990 with a=1 b=29 c=31 and d=2 or if null is allowed, max is 992 a=0 b=1 c=32 d=30

[@asifrashid3054]

If they are different then 1 20 30 12 should give 980

[@WilliamTaylor-h4r]

63/4 ,root mean square, or x^2

[@WDX2738]

How did you proof that the Max occurs when a and d are min? I tried to use Cauchy inequality to solve this but failed.

[@ryanjagpal9457]

@the internet Where the 4 squares?

[@feinfein4444]

@the internet bro you just made me understand this whole bullshit with just a rectangle, but i have some questions Do a,b,c,d have to be different numbers? with your explanation that does mean a and dshould have the lowest value like 1 +2? And then b, c will be like 31 and 29?

[@brainfulify]

@the internet This is why I'm glad to be done with math. A, b, c and d aren't interchangeable in the alphabet, but math gets to be arbitrary.

[@kobebryantlegendary]

I used Cauchy as well with a geometrical approach, but It doesnt work because you have a term missing (a*d). If you had It in there you could verify through Cauchy that, geometrically, for a given quadrileteral with a perimeter P, the biggest area It can have is when all the sides are equal (hence, a Square). If the a*d term was there (and real solutions were allowed), you'd have two answers, one in which a = b = c = d = P/4 (a Square, biggest area), and one in which two of the sides are zero and the other two equal P/2 (that is, when the limit of the two opposite sides tends to zero, hence driving the other two sides tend to their max, which would give a max product of the two, with an optimal max value at the point in which the quadrilateral becomes two coincident lines and stops being a rectangular).

[@jeremy4ags]

awesome video! just want to know if youre from hong kong

[@Forgettings]

I believe there is some ambiguity here in this question; not entirely sure if Cambridge believes that 0 is natural or not but in some areas of math 0 is a natural number and in others it is not. If we assume that zero is in-fact natural Then a=0, b=32, c=31, d=0 would result in bc=992 Which is lager than the answer 991.

[@jean-marcpichand5465]

(31x32+32x0+0x0)=992 31+32+0+0=63 de france from france N* sans 0 N avec 0

[@sadface7457]

My intuition says make a and b approximately equal. Pick c and d such that a and b are maximal.

[@shadowshedinja6124]

Why max out a instead of c?

[@bendaonfire0078]

It waa my first thought too but after a sec or so i saw that C appears twice so its illogical to max out A.

[@sadface7457]

@@shadowshedinja6124 The problem is symmetric c or b both appear twice.

[@shadowshedinja6124]

@@sadface7457 my point is that a doesn't, so a should be smaller

[@samirikar1]

my answer is wrong but i got 16*16+16*16+16*15. where did i go wrong?

[@interesting-g1c]

Funnily enough the solution is actually 992 , because by simply accounting a = d = 0 the remaining b = 31 , c=32 are the correct answer , the test doesn’t say anything about a and d being different from zero

[@fabiofavot2240]

Notice that a,b,c,d are Natural numbers, which conventionally do NOT include zero.

[@gregorysans]

It says natural numbers, which are 1 through infinity, excluding zero.

[@interesting-g1c]

i mean you just need to google natural numbers to see that 0 is included , but ok@@gregorysans

[@wise_math]

Good problem

[@polylab3024]

@mindyourdecisions

[@abhisheksahgamerz4138]

Mind your decision is best

[@C4rnee]

I guessed 991 before writing anything down ☺️ My idea : a and d only show up in "ab+bc+cd" once whereas b and c do twice, so a and d should be minimized (a,b=1), then either b or c = 31, the other = 30.

[@pianissimo7121]

well done, that was literally the approach taken in the video. you have good creativity.

[@Muizur]

Ditto

[@sergiojimenez4595]

same, I solved it while looking at the thumbnail xD it was a pretty easy question though

[@nathancc2526]

Cant u use am or gm here?

[@quo_vadis_europa]

a=19, b=20, c=21, d=3 but only if a != b != c != d otherwise a=31, b=32, c=0, d=0

[@bennethuber7770]

Edit: just ignore this, the first reply pokes a pretty big hole in it due to a silly error on my part Here's a much more intuitive solution. Observe ab + bc + cd = ab + b(c + d). Substitute a' = a, b' = b, c' = c + d. The problem now reduces to max(a'b' + b'c'). We can remove the primes for readability at this point and just say max(ab + bc). Now we can see ab + bc = b(a + c), and a similar substitution reduces the problem to max(ab). Let n = 63, b = n - a. Anyone familiar with calculus can easily prove a(n - a) is maximized at n/2, so we pick a = 32, b = 31. Since the maximum is preserved for more variables, we just set c and d to 0 in the original problem to get our answer. I like this solution because it reveals a fundamental property about breaking up numbers: no matter how many parts you cut them into (since the above substitution can obviously be applied recursively and doesn't depend on the constant), if you sum each part multiplied by it's neighbors, you can never exceed simply dividing the number in half.

[@positivedefinite]

How did you determine that ab + bc + cd = ab + b(c + d)?

[@bennethuber7770]

@@positivedefinite good catch, this is what I get for writing too late at night. It should read ab + c(b + d), but then that breaks the substitution :( . It feels like there should be something here, but I can't figure out how to get it to work right now

[@sabbywins]

Surely the implication of saying a + b + c + d = 63 is that there are four different numbers and, by extension, a =/= d.

[@SloverOfTeuth]

No, it isn't the implication.

[@jeroen6275]

This one was actually pretty easy.

[@adhirajdhar470]

Use basic AM, GM. This is just a cakewalk for JEE aspirants

[@devroopsaha4020]

am-gm kaise lage ga???

[@Armadail_]

Plug eq1 into y=ab+bc+cd, solve the gradient =0 which we get a=d=0, b=c=31.5, finally adjust values to the nearest natural number.

[@MrCarlosmario22]

Que buen ejercicio de optimización.

[@protoroxsinha2451]

Intuitionally finding the proper numbers and checking the products is very difficult u would rather use calculus LaGrange multiplers

[@samibaheru4029]

Since it is natural number we can put the max value 992..*.30+31+2+0=63=a+b+c+d.now do the math.

[@user-km5lt9kg7p]

0 is not natural number

[@samibaheru4029]

Positive integers(whole numbers) 1,2,3,4...and sometimes 0 as well are natural numbers that is why I put two solutions for you.

[@mrslinkydragon9910]

I got 232 A 55 b 4 c 2 d 2 55x4+4×2+2×2

[@bendaonfire0078]

Well.. its not the max so...

[@mitoCoroadoJr]

Lagrange?

[@Speed001]

Neat trick Cambridge, where is this practical?

[@Dharmarajan-ct5ld]

A different approach is given in video solution uploaded now. Kindly consider

[@ffff3c]

1. ab+bc+cd = b(a+c) + cd. Given that equality, if you have d>0, it's ALWAYS better to assign more to b that to d as it increase the total by a for each unite moved from d to b. Thus, d=0 and you're looking for max(b(a+c)) 2. max(b(a+c)) is equalivalent to look for max(x*y) where x+y=63, this is easily realized by x=32,y=31 (or the reverse) N obviously includes 0, but the same logic applies for values in N*: d=1 given the same reasoning above, and since cd is maximized with the largest c, max(b(a+c)) is really calculating max(b(1+c)) where b+c=61, giving b=30, and c=31

[@mrunluckyguy1741]

My dumbass brain thought the numbers must be equal

[@ditran4047]

Isn't the max a=0 b=31 C=32 d=0 since 0 is in N?

[@Vexas345]

Tbh, letting a=d seems like silliness. Else, with the question as written, using d at all just adds fluff and goes against the notion that language (which math is) should be precise and consise to effectively communicate ideas.

[@HichemEllouze18]

It is 992 not 991 When we take a and d 0 b and c will be 31 and 32 (or vice versa) The maximum will be 992

[@horsthorstmann7921]

Only if 0 is a natural number for you.

[@aneeku7519]

Wow wtf ! I got the intuitive technique

[@nice_mf_ngl]

This is a PRMO(india) PYQ which is the first stage of IMO for india, lol I'm seeing people use Lagrange multipliers and shit while actually this was meant for 8-9 graders

[@anandtiwari4080]

Can’t we use am gm inequality

[@cheukyiryanlo943]

road to 10k? nice video w a o

[@jacktank5975]

Shouldn't 'a' not be equal to 'd' as both are different variables altogether ? .... Denoting a,b,c and d are variables.. question says all these numbers are different ie a != b != c != d

[@Muhaiz]

So, why we're learning math actually?

[@nielsverhaeghe9347]

What if a and d are equal to 0, so then b*c=32*31=992, so it’s larger than 991

[@bimarshadhikari5662]

a, b, c and d belong to the set of natural numbers. 0 isn't a natural number.

[@nielsverhaeghe9347]

@@bimarshadhikari5662 oh I always learned that 0 was a natural number. (We learned that you can count with those numbers, and you can have 0 apples).

[@MatsCooper]

Zero together with the natural numbers are called the whole numbers.

[@tharanathakula3588]

If a is not equal d and each has a unique number then what happens?

[@ryanjagpal9457]

idk

[@tonybrowne6737]

Think of a rectangle with sides a+c and b+d. The area is ab+bc+cd+ad. You want to maximize the area of that rectangle and minimize the area ad. The two sides of the rectangle have to add up to 63 so max area is 992. Minimum area of ad is 2 if a and d must be different. So, 990.

[@ryanjagpal9457]

@@tonybrowne6737 Why is it 992 and 990?

[@tonybrowne6737]

@@ryanjagpal9457 I meant the max area of the rectangle (a+c)x(b+d) is 31x32=992. Then you have to take away the ad part. This minimum of this is 1x2=2 if a and d must be different. So the answer to the problem is 990 as you say.

[@ryanjagpal9457]

@@tonybrowne6737 So a+d=2 so you take that away from the equation? if so that is a little confusing

[@sekarganesan]

@letsthincritically - there are few inappropriate comments linking to porn. Pls report or delete if you can.

[@SONUKUMAR-vr2jg]

This was solved earlier by ," Paresh Twalkar"

[@kasyapdharanikota8570]

Very easy question

[@kasyapdharanikota8570]

Just make sure to maximise b and c

[@MaleficeD]

i get 992, a=32, b=31, c=d=0 0 is N

[@yvesdelombaerde5909]

a=d=0, then max is 31x32

[@silasyuen2996]

63/4 times itself time 3 equals 999.1875.

[@BucifalulR]

Natural numbers. 63/4 is not m natural.

[@bendaonfire0078]

2 things. A) 63/4 is not a natural number B) 3*(63/4)^2 is actually 744.1875 So even if it was natural its still not the max

[@quickyummy8120]

Watch @mathematica for more such problems

[@StNick119]

Good video, you have a new subscriber.

[@franciscojavierlara7067]

What does "maximun" and "minimun" mean in this case? Thank you.

[@agoogleuser2942]

Here maximum = the maximum possible value of ab+bc+cd given a+b+c+d=63. For example, when a=1, b=31 or 32 c= 31(if b is 32) or 32 (if b is 31) d=1, the expression ab+bc+cd has the value 991. For no other values of a,b,c and d (satisfying the relation a+b+c+d=63) the expression ab+bc+cd has value greater than 991. Thus the given expression ab+bc+cd has maximum value at a=1, b=31 or 32 c= 31(if b is 32) or 32 (if b is 31) and d=1. Hope it helps 😉.

[@adampaul7905]

a=0 d=0 b=31 c=32 (a+b+c+d=63 and they belong in N ) ab + bc + cd = 31*32 = 992 Your result is si false.

[@Avinashkumar-th6gg]

abcd belongs to natural number

[@midas-holysmoke7642]

I would say 992

[@tllightning12]

Can't you get 992.25 if you do a=0 b=31.5 c=31.5 d=0

[@horsthorstmann7921]

Is 31.5 a natural number?

[@tllightning12]

​@@horsthorstmann7921than what about 32 and 30? That makes 992

[@SloverOfTeuth]

​@@tllightning12Is zero a natural number?

[@tllightning12]

@@SloverOfTeuth tbh idk what they mean by 'natural number'

[@SloverOfTeuth]

@@tllightning12 There's some variation in definitions. Reading the Wikipedia article, I think it traditionally means a positive integer, but some people include zero. It would be clearer if they said "positive integers". This is an interview question. They are interested in admitting people who have the mathematical aptitude to think their way through the problem, not people who know definitions. The Cambridge way of thinking is that you don't get any marks at all for getting the right answer, the marks are for your thinking, that's what they are interested in. They are quite happy to ask questions they know are potentially ambiguous just to see what people do with them. I'm not a mathematician myself, but I've known many many Cambridge mathematicians, I even did the second year computing assignment (Gaussian mesh refinement) for one who didn't click with computers, sitting in the terminal room in DAMTP pretending I was one (ahem).