Alternatively, consider f(a)=a^2(20-a). Since f’(a)=0 at a=40/3, together with other necessary details for the first and/or second derivative tests, we see that max{a*a*b | a+b=20; a, b are positive integers} is the larger of 13*13*7 and 14*14*6, which happens to be the former, being 1183.
I also tested a=1 and a=19. I didn't use the second derivative test in order to show that it was not necessary to test those values. I thought it would be easier just to test them directly.
@@armacham unnecessary to evaluate at a=1 or 19, unless you need to find the (global) min… the derivative tests just help to formally confirm that our intuition stands that f attains its global max (within our range of a) at a=40/3 because it’s the only place where the graph hits a local max and “turns” (first deriv + then 0 then -; second deriv -… either way) - and that f is continuous over the interval
You can also use calculus. Substituting B = 20 - A, A^2*B becomes 20*A^2 - A^3. Taking the derivative of this function yields 40*A - 3*A^2. At the maximum, the rate of change would be 0, so 40*A = 3*A^2. Solving, A = 40÷3 = 13.33. So A = 13 would likely maximize the function A^2*B.
If we say b=20-a, then you can replace that on a²b and you get a cubic, which of course will have some symmetry axis, meaning that if the top vertex in the range [1,19] is a decimal, then the closest integer to that decimal will have bigger value of the function
Local maxima and minima can be checked, and as we would get a cubic equation with maximum value at 40/3~ 13.3333 and as we know any polynomial function is continuous then we would check the integer value near the maxima and the closest integer is 13 to 13.333 so a=13 and b=7 which is (I think) easier then using AM GM inequalities. Thank You!
I don't like your argument why it's sufficient to check 13 and 14. I don't think continuity is a sufficient argument here. Differentiability plus another argument is, though.
@@dyidyirr actually we know how a cubic equation’s graph look and here cubic equation is 20a^2-a^3 and by looking at the expression we can say that it has one up and one down, (you can even compute roots of this equation if you want) by checking it is up (maxima) at 40/3~13.333 and as we know how the cubic graph looks we can pretty much imagine the graph of this equation with one zero at 0 other at 20 on positive x axis and as the up part of the expression(maxima) is like a mountain (I used this term for your understanding) and we have been asked the integer value of a and b then 13 is more close to the ground area where 13.333 lie hence it is input to the cubic equation is highest positive value the graph can provide with a integer input and output. I hope it was clear!
@@jyotiradityajadhav2243 I agree it's correct and you can argue like that. In a beginner course at university I still wouldn't give you full points for that argumentation (that's where you have to be most precise, since they want to teach you to be precise). I would simply argue like this (very verbose): Since p(a) = 20a^2-a^3 is continuously differentiable and p'(a) = 0 if and only if a=0 or a=40/3, the only extrema we can have on the closed set [0,20] are at 0, 40/3, and 20. Between two of these, the function is monotonically increasing or decreasing. Since p(0) = p(20) = 0 < p(40/3) we have a maximum at 40/3, therefore p is increasing between 0 and 40/3 and decreasing between 40/3 and 20. Therefore p(13)>p(n) for all np(n) for all n>14, and therefore it is sufficient to check 13 and 14. But this uses differentiability of p, which you didn't cite in your initial argument. Your second argument does as well, although you don't really say it.
Also we can look at the function f(x) = x^2(20-x). Find the derivative f'(x) = x(40-3x). Extremums are 0 and 40/3. It's easy to figure that 0 is minimum and 40/3 is maximum. The two closest integers to 40/3 are 13 and 14. 169*7 = 1183 and 196*6 = 1176 so a = 13 and b = 7.
Solution is best localized by equating 1st order derivative to zero. That gives real numbers and after that integers in neighborhood can be tried to get precise answer
You can also use lagrange multiplier method. L=(a^2*b+λ(a+b-20). Then Find the first partial derivative with respect to x,y. L'(x) = 0, L'(y)=0, a+b=20. then solve this system of equations.
Thanks for using the AM-GM inequality, I didn't know about it. A faster way for the exam would be to differentiate f(x)=b(20-b)^2 and set it equal to zero.
Without paper or calculator in 1 minute. Subtitute b with 20-a, differentiate, set that to 0, solves as a=40/3. Around this maximum there are 13^2×7=1183, 14^2×6=1176 So answer is 13.
Alternatively, such sums can easily be solved by realising that a^2 b is quasiconcave. For any quasiconcave function, we just need to find marginal rate of substitution i.e. (partial derivative of a^2 b w.r.t b) / (partial derivative w.r.t a) and equalise it slope of the constraint which is 1. We get a=40/3 and b=20/3 which gives the same maximum.
Dude, this is for secondary school students, not uni students. I've heard of partial derivatives but what in the f is a quasiconcave function? Regular calculus applies here, so your method is needlessly complex.
@@jedagelijksebraintraining Sure...it works, but don't _rely_ on it in this kind of situation...you're performing calculus on a discontinuous function, which is a bit of a no-no...
@@bettyswunghole3310 There is an exemple of discontinuous function, where it doesn't work. You are right. If you look in the area of those found extrema, wouldn't you find the correct answer every time? I would test F(12), F(13), F(14), F(15). Can you help me with a good contra-example?
@@chessematics I dunno if this is true for polynomials, but it for sure isn't true for every continuous function. An example is f(x)=2.1/x^2-3/x (yes it's complex, but it's the first example that came to my mind). The minimum is closer to 1, but the value of f(2) is smaller than f(1). Edit: check the function x^3-0.9x. The minimum is closer to 1, but f(0) is smaller than f(1). If it's true, it's true only for quadratics and lower.
@@Noam_.Menashe It is true for linear functions, because it relies on a constant slope. Generally not for polynomials. As an example of a polynomial where Chessematics's line of reasoning does not work take f(x)=30x-x^5. Maximum is at x=sqrt(sqrt(6)) which is around 1.56 The nearest integer is 2. But f(1)=29 and f(2)=28. So the function value at 1 is the biggest!
You don’t need to do this at all, it’s overkill. Geometrically, you just need to consider the side lengths of a cube that maximizes volume. The product of three positive numbers in general is always maximized when the side lengths are as close to equal (equal) as possible. This easily leads to only two choices ie a=13 and b=7 or a=14 and b=6. The explanation of the fact I used earlier needs some calculus to prove, but it is easy enough to have an intuition about especially in an interview. In the case where the domain is any integer number not just naturals the answer is immediate b=20/3.
I actually prefer the geometric method - as mentioned it is intuitive if you figure the maximum area of a 2-D quadrilateral is a square (so both sides are equal) and then apply that thinking to 3-D cuboids.
I think you should check which one between a=13 and a=14 gives a higher a^2b. An easy way to do it without calculating is checking if 13x13x7>14x14x6. 13x13x7/7>14x14/7x6 ; 13x13>14X12 ; 13x13>13x12+12. Checked.
I did it by taking d/da (a^2*(20-a)) = d/da ( 20a^2-a^3) = 40a-3a^2 set this to 0 and you get a^2*b has derivative 0 at a=13.33. Hence the maximum must either be a=13 or a=14, then you calculate a^2*b for a=12,13,14 which gives 1152, 1183, 1176 respectively which confirms the maximum is 1183. My math is very rusty as I am old and retired so I usually figure "forget it" when I see an "Oxford" problem but wow I got one!
I solved it by using Lagrange multiplier of f(a,b)=a2b and g(a,b)=a+b, so 2ab=lambda and a2=lambda, that gives a=sqrt(lambda) and b=sqrt(lambda)/2, thus b=a/2, solving a+b=20 at this condition gives a=40/3 and b=20/3
Consider f(a)=20a^2-a^3' for a=[1,19]. Now f'(a)=40a-3a^2. Solve f'(a)=0, which result a=13.333333. The closest integer will be a=13. Therefore max(a^2b) result when a=13 and b=7.
I did it using calculus, which is as follows. a²b=(20-b)²b= b³-40b²+400b. It's maximum when it's derivative w.r.t b is zero. i.e., 3b²-80b+400=0 i.e., b=20 or b=20/3. We consider b= 7 (a little more than 20/3, as it should be a positive integer). Hurrah ! So easy !
a+b=20 -> b = 20-a, so find the maximum of f(x) = (20-x)*x² on the interval (0,20), just calculate the derivative f'(x)=40x-3x²=0 -> x=0 (minimum) or x=40/3 (maximum) then just compute the value for floor(x)=13 and ceil(x)=14 to get integer values and take the largest, which is for n=13.
your both method could be as right as it is. you just forgot the question. it was not about n, neither x, nor a. as far as we can see, it seemed to be only about the «maximum value of (a²b)» ? including of course the story of finding it.
Or with a little calculus, x = a and b = 20 - x, so a^2 b = 20 x^2 - x^3, now 40 x - 3 x^2 = 0 gives the critical point x = 40 / 3, or a = 13 + 1/3. So it makes sense to test a = 13 and a = 14 and walk away.
Believe it or not, I kinda intuited the answer, after trying a=19, a=18... But I had no idea how to find/prove the answer. So the secret sauce is AM-GM inequality. Thank you!
I used implicit differentiation to find the max of the surface f(a,b)=a^2b. df/da=2ab+a^2(db/da). a+b=20 => db/da=-1. df/da=0=2ab+a^2(-1) => a=2b. a+b=3b=20 => b=20/3 => a=20-20/3. Closest integers are a=13, b=7. Only just realised afterwards I could've just subbed b=20-a into a^2b and differentiated that to get the max...
a^2*b = a^2*(20-a) = -a^3+20a^2 the derivative is -3a^2+40a = -a(3a-40) the extremes will be the roots and 0 is outside of the set 40/3 is the true maximum, limiting it to the natural numbers the closest number is 13
I just substituted b = 20 - a, differentiated and found the stationary points (0 and 40/3), then rounded 40/3 to the nearest integer (13). So a = 13, b = 7.
The second solution is incomplete. Just because the maximum (as given by the AM-GM inequality) occurs at b=20/3, that doesn't mean that among integers the maximum occurs at one of the integers closest to 20/3. One needs to justify that the value of a^2b keeps decreasing from 20/3 in each direction. You can use calculus to check that or if we want to avoid that, compare a^2 b and (a+1)^2 (b-1) to see which one is bigger. Their difference is a^2 - (b-1)(2a+1). If b-1>a/2, then (b-1)(2a+1)>a/2*(2a+1)=a^2+a/2>a^2, so the difference is negative, so (a+1)^2 (b-1)>a^2 b. On the other hand, if b-1
I did it a different way. We are trying to maximise a.a.b where b = 20 - a, so maximise y = a.a.(20 - a) = 20.a.a- a.a.a Pretend for a moment that a is a real number and not an integer. We can find the maximum of y by taking its derivative and equating that to 0. Now, dy/da = 20.2.a - 3.a.a So 0 = 40.a - 3.a.a Hence, a = 0 or 40 = 3.a The first case gives y = 0 and is actually a minimum. The second case gives a = 13.333 Since a must be an integer, 13 is a likely answer, while 14 might be a possibility. For these values, substituting in for a, y = 1183 or y = 1176 , respectively. Clearly 13 gives the maximum. Y is a cubic function of a. It has one maximum an one minimum. It also heads to + infinity for negative values of a, and to - infinitely for positive values. Y = 0 for a = 0 (a double point since y touches the axis there) and y = 0 for a = 20. Of course, you must be careful in this approach of using the continuous case as an approximation for the discrete case. It does not work so well for some other types of problem. But in this one it does work well.
I don't believe that AM-GM gives any information on the growth of the value that's being analyzed, so I don't understand how it can be deduced that the closest integer values to the equality case give the maximum value of the product. It is intuitively true, though, and I think it can be easily seen from calculus.
It is really good channel like a oasis in a desert. Internet enable the chosen ones to access to infinite knowledge but for most of us it is just time consuming.
Lots of inaccuracies here, so be careful if you have to write out your answer. The AM-GM inequality does tell you that equality holds when a/2 = b but it tells you nothing about when a/2 is not equal to b. So the inequality alone is not enough to guarantee that the integer value you are looking for will be the one of the two closest to a/2 = b. And to say that "7 is the closer of 7 and 6 so it must be 7" is not a mathematically valid argument either. Based on knowledge of the the properties of the graph for a^2 * b (expressed in terms of only a or only b), you can safely conclude that b must be either 6 or 7, and you can compare the two.
648... giving 18 and 2, becouse it's logical to assing to a power the maximal value, then take at minimun guarantee at 2 for a major multiplication, not x1, but 2... so when 18*18 = 324, when it doubled, (x2)... reach 648... while 19*19 (19^2) is 361 not better by x1 obvuiosly... leaaving simply 18 = a, and 2 = b... the best combos (18+2=20) and (18^2)*2 = (18*18)*2 = 648 (!)
a + b = 20 b = 20 - a That means: => max(a²(20-a)) Define f(a) := a²(20-a) and just look for the maximum of f. f'(a) = 40a - 3a² => f'(a) = 0 40a - 3a² = 0 a(40 - 3a) = 0 => a_1 = 0 v 40 - 3a = 0. => 40 = 3a a_2 = 40/3 So a_1 or a_2 are the possible maximum values. f''(0) = 40 - 6*0 = 40 > 0 => minimum f''(40/3) = 40 - 6*(40/3) = -40 < 0 => maximum a_2 is the maximum. As result: max(f(a)) = max(a²(20 - a)) = 40/3. But the maximum must be an integer because a and b are integers. That means: Check for a = 13 and a = 14. => f(13) = 1183 => f(14) = 1176 => a = 13 is the right answer.
The answer is correct but the reasoning is not rigorous. f(40/3)>=f(1), f(2), ... , f(19) doesn't imply f(13) or f(14) is the maximum of f(1), f(2), ... , f(19).
Sub a=20-b, we get f(b)=b³-40b²+400b or sub b=20-a, we get g(a)=-a³+20a² a=40/3, b=20/3 is max when solving f'=0 or g'=0 Since I don't want to find unnecessary results, this comes to the art of Maths, f(7)-f(6)>0 f(7)>f(6) (b=7 is max) or g(14)-g(13)g(14) (a=13 is max) max(a²b)=169*7=7*170-7=1183 That's how to compare f(6) & f(7) f(7)-f(6)=(7³-6³)-40(7²-6²)+400(7-6) =(49+42+36)-40(13)+400 =49+42+36-3(40)>0 using a³-b³=(a-b)(a²+ab+b²) & a²-b²=(a+b)(a-b) All the extra steps showing these can be calculated without any calculating devices
In Romania, such a problem is elementary for a third-year high school student. We have b = 20-a. So we need to find max (a ^ 2 (20-a)). Note f (X) = X ^ 2 (20-X). f '(X) = -3X ^ 2 + 40X The sign of f 'is + on (0,13,33) and - on (13,33; infinite) So the maximum value of f is for X= 13.33. But since X is a positive integer, we calculate f (13) and f (14) and we get f is the maximum in X = 13.
@@tobyharrison5468, twice? I only need the points of sign change for df(a)/da, and the direction of the sign changes. for the point of maximum, from + to - with increasing a. Let us suppose that the true point for admission is not just finding the right value of a²b, but telling the right waterproof story about the way we found it?
Easier with concept of derivative Find b in the first equation, substitute the result in the second then derivate and find the zeros for max and min Or you can also use the method of “Hessian matrix of a two-variable function” and then use the first equation to find a and b
By considering the problem as that for maximising the volume of a cube, it can be easily shown that the general solution when a, b, c are positive integers is: * a + b + c = 3n, max(abc) = n^3 * a + b + c = 3n + 1, max(abc) = (n+1).n^2 * a+ b + c = 3n + 2, max(abc) = n(n+1)^2
Honestly I can’t remember how I figured it out, but as a child I taught myself how to approach these value maximizing functions. Now I know exactly how to do it but I can never actually remember why it is that what I’m doing is right.
I like people's approaches but suspect there are assumptions being made about local monotonicity. Is it always true that the integer solution is the integer closest to the real solution?
Not at all For example, consider f(x) = x(x+1.1)^2 and we allow solutions in the integers at least -1. This has a minimum around -0.367, so if this was true we'd expect a minimum at 0. However, f(0) = 0 and f(-1) = -0.01, so the closer integer does not give a smaller value
It says integers, that means negatives too no? If a is a negative integer and b a positive let's say a=-n and b=n+20 (n is a natural) then the sum is 20 but the product a^2b is unbound it grows larger as the n grows larger, I might have something wrong here, I ve been away from math for long
I did this in like 1 min using basic Substitution and Appliying Differentiation, At max value of f(x), f'(x)=0. Applying AM GM is not Nessecary. Well oxford, here I come.
If you differentiate f(x)=-x^3 + 20x^2 to find the stationary points and consider the behaviour of f as x -> +/- infinity, it's clear that x=0 is a local min, x=40/3 a local max, and you don't need to explicitly check x=1 and x=19, though I agree you are doing so indirectly.
Calculus would be easier and more intuitive. There's no way in hell I'd have come up with that oddly specific method during the exam. Even the STEP doesn't require you to know AM/GM. This is the MAT - it only requires AS level maths, as far as I'm aware. (no idea what the American equivalent is, if any)
Much easier to calculate the derivative of a^2(20-a), which gives a maximum at 40/3. a is either 13 or 14. Straight up calculation will give what is larger. In general case; a + b = n; b = (n - a), hence y = a^2(n - a) = na^2 - a^3 dy / da = 2na - 3a^2 eaqualling derivative to 0, 2na - 3a^2 = 0, a = 2n / 3
Hold up you don't known the equality holds when a equals b..how on earth do you reach that conclusion?? You just know the max value 0fna squared b plus at equality so a squared times b equals 1185..
Lagrange be like :: hold my multiplier method 😜😜 f(a,b)=a²b = a²(20-a) ∂f/∂a = a(40-3a)= 0 Therefore either a=0 or a=40/3 So... b= 20-40/3=20/3 Hence Max f = 40/3)² * 20/3 = 32000/27 😜
This solution is smart, but it is not correct. Author proved, that point of maximum among REAL numbers is somewhere between 6 and 7, but this doesn't mean, that maximum among INTEGERS is a closest integer point to maximum among reals. This doesn't mean even, that maximum is 6 or 7, maximum can be any integer point.
@@letsthinkcritically why did you divided a into two for the average why not just a plus b divided by 2..and then do square root instead instead cube root?? It works just as well..
this is incredibly useless and I think it's stupid to sort applicants based on this "test". give me any other higher maths problem, but coming up with a "random" inequality and then working with that out of the blue within 3 minutes is dumb
For y up to 2000 we can use python: b = 1 for i in range(2001): a = i*i*(2000-i) if a > max(19,b): # the smallest value is (1^2)* 19 b = a print(b) 1185184963
![[DRAGON BALL LEGENDS REVEALS ＆ STUFF] LEGENDS FESTIVAL 2024 Special Edition Part 2](http://i.ytimg.com/vi/af0phKxitX8/mqdefault.jpg)
Alternatively, consider f(a)=a^2(20-a). Since f’(a)=0 at a=40/3, together with other necessary details for the first and/or second derivative tests, we see that max{a*a*b | a+b=20; a, b are positive integers} is the larger of 13*13*7 and 14*14*6, which happens to be the former, being 1183.
@@geneyoungdho tu
Exactly.
That’s how I thought about the problem too lol.
I also tested a=1 and a=19. I didn't use the second derivative test in order to show that it was not necessary to test those values. I thought it would be easier just to test them directly.
@@armacham unnecessary to evaluate at a=1 or 19, unless you need to find the (global) min…
the derivative tests just help to formally confirm that our intuition stands that f attains its global max (within our range of a) at a=40/3 because it’s the only place where the graph hits a local max and “turns” (first deriv + then 0 then -; second deriv -… either way) - and that f is continuous over the interval
You can also use calculus. Substituting B = 20 - A, A^2*B becomes 20*A^2 - A^3. Taking the derivative of this function yields 40*A - 3*A^2. At the maximum, the rate of change would be 0, so 40*A = 3*A^2. Solving, A = 40÷3 = 13.33. So A = 13 would likely maximize the function A^2*B.
Incredibly chad comment
You also have to check A = 14 too
@@JoeMama-xq8bf nah as it rounds closer to 13
incredible way of thinking bro
@@JoeMama-xq8bf Good point, doing the calc for 13 and 14 is the final step I left out.
Just because 20/3 is closer to 7 doesn't mean the 7 is the answer, you still need to check both b=6 and b=7.
I agree: the fact that a is divided by 2 and appears more than once in the sum might have messed this reasoning of "closer" up.
Correct check both
nope, the closer the variables are, the bigger the GM
If we say b=20-a, then you can replace that on a²b and you get a cubic, which of course will have some symmetry axis, meaning that if the top vertex in the range [1,19] is a decimal, then the closest integer to that decimal will have bigger value of the function
@@eterty8335 there are ways to prove it’s 7, not shown in the video and that is clearly not the method they use
Local maxima and minima can be checked, and as we would get a cubic equation with maximum value at 40/3~ 13.3333 and as we know any polynomial function is continuous then we would check the integer value near the maxima and the closest integer is 13 to 13.333 so a=13 and b=7 which is (I think) easier then using AM GM inequalities. Thank You!
I think you mean 40/3
@@liamtolkkinen5025 yes I am sorry for the mistake, thanks for correcting!
I don't like your argument why it's sufficient to check 13 and 14. I don't think continuity is a sufficient argument here. Differentiability plus another argument is, though.
@@dyidyirr actually we know how a cubic equation’s graph look and here cubic equation is 20a^2-a^3 and by looking at the expression we can say that it has one up and one down, (you can even compute roots of this equation if you want) by checking it is up (maxima) at 40/3~13.333 and as we know how the cubic graph looks we can pretty much imagine the graph of this equation with one zero at 0 other at 20 on positive x axis and as the up part of the expression(maxima) is like a mountain (I used this term for your understanding) and we have been asked the integer value of a and b then 13 is more close to the ground area where 13.333 lie hence it is input to the cubic equation is highest positive value the graph can provide with a integer input and output. I hope it was clear!
@@jyotiradityajadhav2243 I agree it's correct and you can argue like that. In a beginner course at university I still wouldn't give you full points for that argumentation (that's where you have to be most precise, since they want to teach you to be precise). I would simply argue like this (very verbose): Since p(a) = 20a^2-a^3 is continuously differentiable and p'(a) = 0 if and only if a=0 or a=40/3, the only extrema we can have on the closed set [0,20] are at 0, 40/3, and 20. Between two of these, the function is monotonically increasing or decreasing. Since p(0) = p(20) = 0 < p(40/3) we have a maximum at 40/3, therefore p is increasing between 0 and 40/3 and decreasing between 40/3 and 20. Therefore p(13)>p(n) for all np(n) for all n>14, and therefore it is sufficient to check 13 and 14. But this uses differentiability of p, which you didn't cite in your initial argument. Your second argument does as well, although you don't really say it.
Also we can look at the function f(x) = x^2(20-x). Find the derivative f'(x) = x(40-3x). Extremums are 0 and 40/3. It's easy to figure that 0 is minimum and 40/3 is maximum. The two closest integers to 40/3 are 13 and 14. 169*7 = 1183 and 196*6 = 1176 so a = 13 and b = 7.
Solution is best localized by equating 1st order derivative to zero. That gives real numbers and after that integers in neighborhood can be tried to get precise answer
This has two variables, so technically you will find a gradient.... Multiple critical points must be evaluated.
@@vesselofmercy6988 b = 20-a, so it's really just one variable.
You can also use lagrange multiplier method. L=(a^2*b+λ(a+b-20). Then Find the first partial derivative with respect to x,y.
L'(x) = 0,
L'(y)=0,
a+b=20. then solve this system of equations.
that takes me back!!!
Thanks for using the AM-GM inequality, I didn't know about it. A faster way for the exam would be to differentiate f(x)=b(20-b)^2 and set it equal to zero.
Without paper or calculator in 1 minute. Subtitute b with 20-a, differentiate, set that to 0, solves as a=40/3. Around this maximum there are 13^2×7=1183, 14^2×6=1176 So answer is 13.
Exactly what I did. I was afraid I was the only one who did it the easy way.
Alternatively, such sums can easily be solved by realising that a^2 b is quasiconcave. For any quasiconcave function, we just need to find marginal rate of substitution i.e. (partial derivative of a^2 b w.r.t b) / (partial derivative w.r.t a) and equalise it slope of the constraint which is 1. We get a=40/3 and b=20/3 which gives the same maximum.
Ppl applying to study for maths won't know this unless they covered partial derivatives and quasiconcave functions (which I've never heard until now)
Dude, this is for secondary school students, not uni students.
I've heard of partial derivatives but what in the f is a quasiconcave function?
Regular calculus applies here, so your method is needlessly complex.
F(a)=a^2.(20-a)
F(a) = 20a^2 - a^3
First derivative...
F'(a)= 40a-3a^2 = 0 for extremum
So a = 0 makes a mimimum.
So a = 40/3 makes a maximum.
This only applies "by lucky coincidence" when dealing with integral values.
@@bettyswunghole3310 this is the way I learned to get extremums.
@@jedagelijksebraintraining Sure...it works, but don't _rely_ on it in this kind of situation...you're performing calculus on a discontinuous function, which is a bit of a no-no...
@@bettyswunghole3310 There is an exemple of discontinuous function, where it doesn't work. You are right. If you look in the area of those found extrema, wouldn't you find the correct answer every time? I would test F(12), F(13), F(14), F(15).
Can you help me with a good contra-example?
Great! I tried turning it into a function x^2*(20-x) and finding the nearest integer closest to its maximum value, i.e. when x=40/3.
It is good to know the graph of that polinom, and we see that f'(x) =0 for x=40/3 so we need to check a=13 and a=14
@@זאבגלברד don't need to check. 13 is closer to 40/3 than 14. That's it
@@chessematics I dunno if this is true for polynomials, but it for sure isn't true for every continuous function. An example is f(x)=2.1/x^2-3/x (yes it's complex, but it's the first example that came to my mind). The minimum is closer to 1, but the value of f(2) is smaller than f(1).
Edit: check the function x^3-0.9x. The minimum is closer to 1, but f(0) is smaller than f(1). If it's true, it's true only for quadratics and lower.
@@Noam_.Menashe It is true for linear functions, because it relies on a constant slope. Generally not for polynomials.
As an example of a polynomial where Chessematics's line of reasoning does not work take f(x)=30x-x^5. Maximum is at x=sqrt(sqrt(6)) which is around 1.56
The nearest integer is 2. But f(1)=29 and f(2)=28. So the function value at 1 is the biggest!
@@koenth2359 linear functions don't have extrema....
You don’t need to do this at all, it’s overkill. Geometrically, you just need to consider the side lengths of a cube that maximizes volume. The product of three positive numbers in general is always maximized when the side lengths are as close to equal (equal) as possible. This easily leads to only two choices ie a=13 and b=7 or a=14 and b=6. The explanation of the fact I used earlier needs some calculus to prove, but it is easy enough to have an intuition about especially in an interview. In the case where the domain is any integer number not just naturals the answer is immediate b=20/3.
I actually prefer the geometric method - as mentioned it is intuitive if you figure the maximum area of a 2-D quadrilateral is a square (so both sides are equal) and then apply that thinking to 3-D cuboids.
I think you should check which one between a=13 and a=14 gives a higher a^2b. An easy way to do it without calculating is checking if 13x13x7>14x14x6.
13x13x7/7>14x14/7x6 ; 13x13>14X12 ; 13x13>13x12+12. Checked.
I did it by taking d/da (a^2*(20-a)) = d/da ( 20a^2-a^3) = 40a-3a^2 set this to 0 and you get a^2*b has derivative 0 at a=13.33. Hence the maximum must either be a=13 or a=14, then you calculate a^2*b for a=12,13,14 which gives 1152, 1183, 1176 respectively which confirms the maximum is 1183. My math is very rusty as I am old and retired so I usually figure "forget it" when I see an "Oxford" problem but wow I got one!
I just used calculus to find that the local maximum of f(a) = a^2*(20-a) is at a = 40/3, then tested the integers on either side of 40/3.
Nice exact same process
Explain pls
@@Daniel_k729 take the derivative via product rule and then solve for when it’s 0 for critical points
@@thewitchking2556 I am 15 so idk but I will research on what that is
@@Daniel_k729 It's very early calculus, which is basically what you need to study optimization, maximums of functions, and minimums of functions
I solved it by using Lagrange multiplier of f(a,b)=a2b and g(a,b)=a+b, so 2ab=lambda and a2=lambda, that gives a=sqrt(lambda) and b=sqrt(lambda)/2, thus b=a/2, solving a+b=20 at this condition gives a=40/3 and b=20/3
This is a very easy problem, what I did was i created a cubic and solved for when the derivative is 0. You’ll get 40/3 and 20/3 for a and b
Interestingly, if a and b are not integers the maximum is when a = 40/3 and b = 20/3, which gives a²b = 1185.1851851851851852...
I got the same answer using derivative, but not sure if right or wrong
I just optimized it. If decimals are allowed, a = 13.33 and b = 6.67
If only natural numbers are allowed, simply just round them up to 13 and 7.
well done, no
This method is quite intuitive and good for comprehension.....for we can do this easily and quickly by using differentiation
Consider f(a)=20a^2-a^3' for a=[1,19].
Now f'(a)=40a-3a^2.
Solve f'(a)=0, which result a=13.333333. The closest integer will be a=13.
Therefore max(a^2b) result when a=13 and b=7.
I did it using calculus, which is as follows. a²b=(20-b)²b= b³-40b²+400b. It's maximum when it's derivative w.r.t b is zero. i.e., 3b²-80b+400=0 i.e., b=20 or b=20/3. We consider b= 7 (a little more than 20/3, as it should be a positive integer). Hurrah ! So easy !
a+b=20 -> b = 20-a, so find the maximum of f(x) = (20-x)*x² on the interval (0,20), just calculate the derivative f'(x)=40x-3x²=0 -> x=0 (minimum) or x=40/3 (maximum) then just compute the value for floor(x)=13 and ceil(x)=14 to get integer values and take the largest, which is for n=13.
My exact method!
your both method could be as right as it is.
you just forgot the question. it was not about n, neither x, nor a.
as far as we can see, it seemed to be only about the
«maximum value of (a²b)» ?
including of course the story of finding it.
I noticed a mistake at 1:29:
5^2 x 15 = 375, not 325 as shown in the video
Or with a little calculus, x = a and b = 20 - x, so a^2 b = 20 x^2 - x^3, now 40 x - 3 x^2 = 0 gives the critical point x = 40 / 3, or a = 13 + 1/3. So it makes sense to test a = 13 and a = 14 and walk away.
OMG...I love your video so much....so hard and challenging but all are theorical thinking problem....I learn so much from your video
Just did it through lagrange multipliers pretty quickly.
Believe it or not, I kinda intuited the answer, after trying a=19, a=18... But I had no idea how to find/prove the answer. So the secret sauce is AM-GM inequality. Thank you!
I used implicit differentiation to find the max of the surface f(a,b)=a^2b.
df/da=2ab+a^2(db/da).
a+b=20 => db/da=-1.
df/da=0=2ab+a^2(-1) => a=2b.
a+b=3b=20 => b=20/3 => a=20-20/3.
Closest integers are a=13, b=7.
Only just realised afterwards I could've just subbed b=20-a into a^2b and differentiated that to get the max...
a^2*b = a^2*(20-a) = -a^3+20a^2
the derivative is -3a^2+40a = -a(3a-40)
the extremes will be the roots and 0 is outside of the set
40/3 is the true maximum, limiting it to the natural numbers the closest number is 13
a=20-b max(a^2 b) is double derivative of b(20-b)^2 -> 6b-80=0 b is 13 closest integer
I just substituted b = 20 - a, differentiated and found the stationary points (0 and 40/3), then rounded 40/3 to the nearest integer (13). So a = 13, b = 7.
The second solution is incomplete. Just because the maximum (as given by the AM-GM inequality) occurs at b=20/3, that doesn't mean that among integers the maximum occurs at one of the integers closest to 20/3. One needs to justify that the value of a^2b keeps decreasing from 20/3 in each direction. You can use calculus to check that or if we want to avoid that, compare a^2 b and (a+1)^2 (b-1) to see which one is bigger. Their difference is a^2 - (b-1)(2a+1). If b-1>a/2, then (b-1)(2a+1)>a/2*(2a+1)=a^2+a/2>a^2, so the difference is negative, so (a+1)^2 (b-1)>a^2 b. On the other hand, if b-1
I did the longer function b x (20-b)^2, after finding extrema, I got 20/3, that is 6.66667, so when rounded we get b = 7 a = 13
same solution and AM-GM inequity is one of my favorite formulas.
I did it a different way.
We are trying to maximise a.a.b where b = 20 - a, so maximise y = a.a.(20 - a) = 20.a.a- a.a.a
Pretend for a moment that a is a real number and not an integer.
We can find the maximum of y by taking its derivative and equating that to 0.
Now, dy/da = 20.2.a - 3.a.a
So 0 = 40.a - 3.a.a
Hence, a = 0 or 40 = 3.a
The first case gives y = 0 and is actually a minimum.
The second case gives a = 13.333
Since a must be an integer, 13 is a likely answer, while 14 might be a possibility.
For these values, substituting in for a, y = 1183 or y = 1176 , respectively. Clearly 13 gives the maximum.
Y is a cubic function of a. It has one maximum an one minimum.
It also heads to + infinity for negative values of a, and to - infinitely for positive values.
Y = 0 for a = 0 (a double point since y touches the axis there) and y = 0 for a = 20.
Of course, you must be careful in this approach of using the continuous case as an approximation for the discrete case.
It does not work so well for some other types of problem. But in this one it does work well.
I don't believe that AM-GM gives any information on the growth of the value that's being analyzed, so I don't understand how it can be deduced that the closest integer values to the equality case give the maximum value of the product. It is intuitively true, though, and I think it can be easily seen from calculus.
It is really good channel like a oasis in a desert.
Internet enable the chosen ones to access to infinite knowledge but for most of us it is just time consuming.
Lots of inaccuracies here, so be careful if you have to write out your answer. The AM-GM inequality does tell you that equality holds when a/2 = b but it tells you nothing about when a/2 is not equal to b. So the inequality alone is not enough to guarantee that the integer value you are looking for will be the one of the two closest to a/2 = b. And to say that "7 is the closer of 7 and 6 so it must be 7" is not a mathematically valid argument either. Based on knowledge of the the properties of the graph for a^2 * b (expressed in terms of only a or only b), you can safely conclude that b must be either 6 or 7, and you can compare the two.
For what vale of a is the a^2 /(20-a) maximum, a < 20.
648...
giving 18 and 2,
becouse it's logical to assing to a power
the maximal value, then take at minimun guarantee at 2 for a major multiplication, not x1, but 2... so when 18*18 = 324, when it doubled, (x2)... reach 648...
while 19*19 (19^2) is 361 not better by x1 obvuiosly... leaaving simply 18 = a, and 2 = b... the best combos (18+2=20) and (18^2)*2 = (18*18)*2 = 648 (!)
Jo's palaria! Super rezolvare! Bravo! Remarcabila demonstratie! 😍👀🙄👍👌😊🙏🙌👌🏆🇷🇴🙏💯🏆
a + b = 20 b = 20 - a
That means:
=> max(a²(20-a))
Define f(a) := a²(20-a) and just look for the maximum of f.
f'(a) = 40a - 3a²
=> f'(a) = 0 40a - 3a² = 0
a(40 - 3a) = 0
=> a_1 = 0 v 40 - 3a = 0.
=> 40 = 3a a_2 = 40/3
So a_1 or a_2 are the possible maximum values.
f''(0) = 40 - 6*0 = 40 > 0 => minimum
f''(40/3) = 40 - 6*(40/3) = -40 < 0 => maximum
a_2 is the maximum.
As result:
max(f(a)) = max(a²(20 - a)) = 40/3.
But the maximum must be an integer because a and b are integers.
That means:
Check for a = 13 and a = 14.
=> f(13) = 1183
=> f(14) = 1176
=> a = 13 is the right answer.
The answer is correct but the reasoning is not rigorous. f(40/3)>=f(1), f(2), ... , f(19) doesn't imply f(13) or f(14) is the maximum of f(1), f(2), ... , f(19).
@@nychan2939 I didn't notice it at the moment. My bad
Sub a=20-b, we get f(b)=b³-40b²+400b
or sub b=20-a, we get g(a)=-a³+20a²
a=40/3, b=20/3 is max when solving f'=0 or g'=0
Since I don't want to find unnecessary results, this comes to the art of Maths,
f(7)-f(6)>0
f(7)>f(6) (b=7 is max)
or g(14)-g(13)g(14) (a=13 is max)
max(a²b)=169*7=7*170-7=1183
That's how to compare f(6) & f(7)
f(7)-f(6)=(7³-6³)-40(7²-6²)+400(7-6)
=(49+42+36)-40(13)+400
=49+42+36-3(40)>0
using a³-b³=(a-b)(a²+ab+b²) & a²-b²=(a+b)(a-b)
All the extra steps showing these can be calculated without any calculating devices
In Romania, such a problem is elementary for a third-year high school student.
We have b = 20-a. So we need to find max (a ^ 2 (20-a)).
Note f (X) = X ^ 2 (20-X).
f '(X) = -3X ^ 2 + 40X
The sign of f 'is + on (0,13,33) and - on (13,33; infinite)
So the maximum value of f is for X= 13.33. But since X is a positive integer, we calculate f (13) and f (14) and we get f is the maximum in X = 13.
yeah thats what i got too when I differentiated it twice
@@tobyharrison5468,
twice?
I only need the points of sign change for df(a)/da, and the direction of the sign changes.
for the point of maximum, from + to - with increasing a.
Let us suppose that the true point for admission is not just finding the right value of a²b, but telling the right waterproof story about the way we found it?
@@keescanalfp5143 yeah i used the second derivative to find the local maximum for and b
thanks for sharing
Easier with concept of derivative
Find b in the first equation, substitute the result in the second then derivate and find the zeros for max and min
Or you can also use the method of “Hessian matrix of a two-variable function” and then use the first equation to find a and b
I just solved it my head by going through possible combinations.
By considering the problem as that for maximising the volume of a cube, it can be easily shown that the general solution when a, b, c are positive integers is:
* a + b + c = 3n, max(abc) = n^3
* a + b + c = 3n + 1, max(abc) = (n+1).n^2
* a+ b + c = 3n + 2, max(abc) = n(n+1)^2
Honestly I can’t remember how I figured it out, but as a child I taught myself how to approach these value maximizing functions. Now I know exactly how to do it but I can never actually remember why it is that what I’m doing is right.
🙏🏻 proffessor sir, answer,1183
Where a=13,b=7
Just a typo correction: 5^2*15=375 not 325!
It just similar as cambridge interview question(a+b+c+d=63, max(ab+bc+cd)? )
Yeah AM GM inequality was the first thing that came to mind, did it mentally till 32000/27, didn't bother to check with the correct number tho
Hi, algebra god, I find that for those geometric guys, this question using cubic graph to solve is just so simple
Epic trollge moment
I like people's approaches but suspect there are assumptions being made about local monotonicity. Is it always true that the integer solution is the integer closest to the real solution?
Not at all
For example, consider f(x) = x(x+1.1)^2 and we allow solutions in the integers at least -1. This has a minimum around -0.367, so if this was true we'd expect a minimum at 0. However, f(0) = 0 and f(-1) = -0.01, so the closer integer does not give a smaller value
Did it in my head, going by a (roughly) binary chop sequence.
Why use linear algebra when you got calculus?👀
It says integers, that means negatives too no? If a is a negative integer and b a positive let's say a=-n and b=n+20 (n is a natural) then the sum is 20 but the product a^2b is unbound it grows larger as the n grows larger, I might have something wrong here, I ve been away from math for long
Maybe that's trivial, but how do you know that the a^2b
What he means is that the sides are actually equal when b=a/2
If it wasn not given that a,b are positive integer numbers, would this problem be solved?
convexity !
3 × 7 = 21
a bit smaller than 7 , will do.
I did this in like 1 min using basic Substitution and Appliying Differentiation, At max value of f(x), f'(x)=0. Applying AM GM is not Nessecary. Well oxford, here I come.
i felt AM/GM does require more thinking, taking derivative is like using formula to solve general quadratic equation
At what age should I be able to solve this
If you're using derivatives , you MUST also check a=1 and a=19;)
If you differentiate f(x)=-x^3 + 20x^2 to find the stationary points and consider the behaviour of f as x -> +/- infinity, it's clear that x=0 is a local min, x=40/3 a local max, and you don't need to explicitly check x=1 and x=19, though I agree you are doing so indirectly.
my brain kinda blows up every time i see AM-GM inequality in the solution cuz i keep forgetting to use it
Rearrangement + Substitution + Differentiation (1st & 2nd Order) to check stationary point(s) for maximum? 😂😂
Calculus would be easier and more intuitive. There's no way in hell I'd have come up with that oddly specific method during the exam. Even the STEP doesn't require you to know AM/GM. This is the MAT - it only requires AS level maths, as far as I'm aware. (no idea what the American equivalent is, if any)
Much easier to calculate the derivative of a^2(20-a), which gives a maximum at 40/3. a is either 13 or 14. Straight up calculation will give what is larger.
In general case;
a + b = n; b = (n - a), hence y = a^2(n - a) = na^2 - a^3
dy / da = 2na - 3a^2
eaqualling derivative to 0, 2na - 3a^2 = 0, a = 2n / 3
Hold up you don't known the equality holds when a equals b..how on earth do you reach that conclusion?? You just know the max value 0fna squared b plus at equality so a squared times b equals 1185..
nice
I made it with calculus in what felt like about three minutes (but I didn't keep the time...)
I swear, the lengths people are willing to go to avoid doing calculus
You published this on April 1st? And it’s actually true? I’m still looking for the joke… ;)
We can easily do it by AM>=GM
Lagrange be like :: hold my multiplier method 😜😜
f(a,b)=a²b = a²(20-a)
∂f/∂a = a(40-3a)= 0
Therefore either a=0 or a=40/3
So... b= 20-40/3=20/3
Hence
Max f = 40/3)² * 20/3 = 32000/27 😜
Using calculus after a substitution is easier.
3 mn is too much max(a²b)=1183.
Ah! Messed up at the last
This solution is smart, but it is not correct. Author proved, that point of maximum among REAL numbers is somewhere between 6 and 7, but this doesn't mean, that maximum among INTEGERS is a closest integer point to maximum among reals. This doesn't mean even, that maximum is 6 or 7, maximum can be any integer point.
If "a" was "5"and "b" "-5". I.. would think.. lol..
Anyone else who chose a=19 intuitively?
My grandma solved it by differentiation.Can my grandma get admission?
This is easier with precalculus
at 1:28 you mistakenly multiplied by 13 rather than 15.
Thank you for pointing that out. The product should be 375.
@@letsthinkcritically why did you divided a into two for the average why not just a plus b divided by 2..and then do square root instead instead cube root?? It works just as well..
Calculus can solve it
guys,we do that math in 17 years old kids in greece.get serious. it's a simple max function problem.come on...7 min video for 30 seconds solution
Bro it's just calculus, I just solved it within 3 minutes
set a=20-b then basic calculus lol
Not even calculus, you can actually find a=13 using only algebra and the quadratic formula without any case to check
I just spammed derivatives
११८३
this is incredibly useless and I think it's stupid to sort applicants based on this "test". give me any other higher maths problem, but coming up with a "random" inequality and then working with that out of the blue within 3 minutes is dumb
Take a=10,b=10 result=1000
For y up to 2000 we can use python:
b = 1
for i in range(2001):
a = i*i*(2000-i)
if a > max(19,b): # the smallest value is (1^2)* 19
b = a
print(b)
1185184963