How Cauchy would find the maximum of sqrt(x)+sqrt(y)

[@blackpenredpen]

If x+y=8, then find the max of x^y Answer here: kzbin.info/www/bejne/sJWke4ufoZKBrKM

[@leonardobarrera2816]

It works for complex numbers?

[@keescanalfp5143]

​@@leonardobarrera2816, well we don't think so . among complex numbers you have no >,

[@leonardobarrera2816]

@@keescanalfp5143idk Well, I didn’t knew about those functions I need to study them But, plz, if you know something occurs let me know It is really fun!!!!

[@jasonhakim832]

please i found an integral in my computation help me solving it : the integral of 1/(arcos(x)(1-x))

[@evancisolomon2930]

I've done by using 2nd derivative test

[@nikoskonstantinou3681]

Well, if you substitute the variable x and y for u² and w², then the question is: Given u²+2w²=4, find maximum of u+w. The first equation is an elipse and the second, supposing that the maximum is a number "m", can be written as a line: u+w=m => w=-u+m, which obviously has a negative slope. The m is maximum only when the line and the elipse are tangent with each other (if they have 2 common points, then the m can be increased until they have one common point, e.g. they are tagent). Since u and w are positive numbers, and the slope is negative we only care for the case of the 1st quarter for the elipse. Solving the system for the tagent of the elipse, with slope -1, produces the answer m=√6. I like this approach since it is more visual and does not require a prior knowledge of some commont (or not) inequalities

[@zay_dog1267]

to add a shortcut to your approach: we can parameterise the ellipse with u=2cos(t) and w=sqrt(2)*sin(t), so z+w=2cos(t)+sqrt(2)*sin(t) which by harmonic form is equal to sqrt(6)*cos(t-phi) as the amplitudes add pythagoreanly, so the max is just sqrt(6) - barely any number crunching needed :)

[@luismuller6505]

I did it exactly this way as well.

[@kappasphere]

My solution using Lagrange multipliers: (I noticed it's total overkill) F(x, y, λ)=x^½+y^½+λ(x+2y-4) dF/dx=1/(2 x^½) + λ=0 dF/dy=1/(2 y^½) + 2λ=0 dF/dλ=x+2y-4=0 Take II.-2I.: 1/(2 y^½) = 1/x^½ 4 y = x Insert into III.: 6y=4 y=2/3 x=8/3 To determine if this is a maximum, compare to the boundaries at x=0 or y=0: x=0,y=2,x^½+y^½=sqrt(2) x=4,y=0,x^½+y^½=2 x=8/3,y=2/3,x^½+y^½=3 sqrt(⅔)=sqrt(6) sqrt(6) > sqrt(4) = 2 > sqrt(2) So the maximum is sqrt(6) with x=8/3 and y=2/3

[@Name-ej4ex]

I was hoping that he would use this.

[@VenkataB123]

Even I used Lagrange multipliers to solve this lol! Didn't really think it was overkill, as the lambda gets eliminated pretty easily. I've seen way worse problems where there's so many possible values of lambda and many relations between the variables that I lose track of them💀

[@suyashmisra7406]

Anf i thought i was cheating by using derivatives

[@abrahamplasencia2611]

wdym when you wrote sqrt(6) = 2 sqrt(3) > 2 > sqrt(2), bro sqrt(6) is not equal to 2 sqrt(3), you could have just written that 6>2 so sqrt( 6) > square root (2)

[@kappasphere]

@@abrahamplasencia2611 I probably thought 6=4*3 when it was only 2*3. I'll edit the comment to have a proper comparison, thanks.

[@jimmy_4]

thank you so much, i was having trouble understanding this inequality for so many months..this really helped..

[@agentbinodbollywoodwale6656]

Here is my solution for the problem using purely coordinate geometry. This problem can be written as: "Maximize sqrt(x) +sqrt(y) subject to condition x + 2y = 4, where x and y are real numbers." So, if we take p = sqrt(x) and q = sqrt(y), and change the coordinate system of the above problem from XY plane to PQ plane, the problem can be translated as: "Maximize p + q subject to condition p ^ 2 + 2(q ^ 2) = 4, where p and q are real numbers greater than or equal to 0." This changes the function we want to maximize to a linear function and the conditions suggest that the point (p, q) is a point on the ellipse p ^ 2 + 2(q ^ 2) = 4 in the first quadrant. We want to maximize linear function p + q for such points. Suppose for any such point (p, q), p + q = r. Then we can say that the point (p, q) is the point of intersection of the straight line p + q = r and the ellipse p ^ 2 + 2(q ^ 2) = 4 in the first quadrant. If we draw the diagram for it, then we can see that for larger values of r, the straight line p + q = r seems to move away from the centre of the ellipse and hence after one point not intersecting the ellipse in the first quadrant. So, the maximum value of r for which the intersection between the straight line and the ellipse can be maintained is when p + q = r becomes tangent to p ^ 2 + 2(q ^ 2) = 4. So, for the point (p, q) on the ellipse, the equation of tangent at this point will be: p * P + 2 * q * Q = 4 which should be the same as 1 * P + 1 * Q = r, P and Q are taken as variables here. Comparing these we get, p / 1 = 2 * q / 1 = 4 / r which implies that p = 4 / r and q = 2 / r. As point (p, q) lies on the line P + Q = r, therefore, (4 / r) + (2 / r) = r which implies that 6 / r = r which further implies that r ^ 2 = 6 and hence r = sqrt(6) as P and Q are positive, r MUST be positive. Hence the maximum value of p + q = sqrt(6) which is sqrt(x) + sqrt(y) = sqrt(6).

[@caiolira09]

This teacher is the Bob Ross of mathematics. Always a pleasure watching his videos!

[@chsh9686]

Let f be defined by f(y) = sqrt(y) + sqrt (4-2y). f is defined as long as 4-2y >= 0, meaning y= 0, which means f is defined on [0;2]. Let's find f' : f'(y) = 1/2/sqrt(y) - 1/sqrt(4-2y) Let's now solve f'(y)=0 It means 1/(2*sqrt(y))= 1/sqrt(4-2y) 2*sqrt(y)=sqrt(4-2y) 4y=4-2y y=2/3 Because it is obvious that f increases then decreases (draw the function if needed), f(2/3) will be the max of the function (called M) M = sqrt(2/3) + sqrt(8/3) M = (sqrt(2) + 2*sqrt(2))/sqrt(3) M = 3*sqrt(2)/sqrt(3) M = sqrt(3)*sqrt(2) M = sqrt(6)

[@sebmata135]

Very nice. I used Lagrange Multipliers myself. Was pretty easy since letting F(x,y)=sqx+sqy gave gradF=. Then letting the constraint G(x,y)=x+2y=4 gave gradG=. Writing gradF=c*gradG gave the equations 1/sqx=c*1 and 1/sqy=c*2, then subbing gave 1/sqy=2/sqx. Cross multiplying gave sqx=2sqy, and squaring gave x=4y. Plugging this into the constraint gave 6y=4 so that y = 2/3, then x=4*2/3=8/3. Then F(8/3,2/3)=sq6. Obviously you also check that F(0,0)=0 isn't the max since that's the only point at which F isn't differentiable.

[@jubinsoni4694]

I solved the same way, as soon as maxima was asked, question was screaming to do dy/dx = 0, thanks for sharing 👍

[@ShahzadaMMunir]

I do not understand how the sqrt(8/3) was figured out in the last section, could you explain by any chance?

[@chsh9686]

@@ShahzadaMMunir I just found before that the maximum was for y=2/3 so I just replaced y by 2/3 in the definition of f. f(y) = sqrt(y) + sqrt (4-2y) and with y = 2/3, it gives f(y) = sqrt(2/3) + sqrt(4-2*2/3) And of course 4-2*2/3 = 8/3

[@ShahzadaMMunir]

@@chsh9686 Truly appreciated. Thank you very much

[@spectrumjitters4672]

I know no one cares, but I was able to solve this in my head by looking at the thumbnail in a couple minutes. I used calculus optimization

[@Ninja20704]

The general case of the inequality is [∑(a_i * b_i)]^2

[@El0melette]

Actually the general case is for any Hilbert space (H, p( , )), we have, for all x,y in H, |p(x,y)|

[@matheusjahnke8643]

@@El0melette inner product can be as simple as multiplication between scalars(which the equality always holds[there's something for complex numbers... won't go into that])... A dot product between two vectors; Or.... for random variables X and Y with mean 0... their covariance... E[XY] (the expected value of X times Y... which can be a sum for discrete variables... or a integral for continuous variables). Inner product it's just an operation with some properties

[@El0melette]

@@matheusjahnke8643 Yes, of course and that's the reason why I say "the general case". What I do have to correct is that it is not necessary for it to be Hilbert, a space with an inner product and norm induced by it is enough.

[@yoyoezzijr]

Beautiful solution. Love your content.

[@blackpenredpen]

Thank you! Cheers!

[@jyotsanabenpanchal7271]

Plz make a video on stolz Cesaro theorem.😢​@@blackpenredpen

[@milos_radovanovic]

Even if we take the C-S inequality as a given, how do we prove that this method will always produce a true minimum or maximum, not just a lower or upper bound?

[@sclearDevelopment]

In the video, he mentioned when the equality is found (a/c=b/d). This method gives an upper bound, so I think it will take a little more effort to change the system to find the minimum with it. He bounded it from below by 0, but that isn't the true minimum. I believe the true minimum is sqrt 2 (x=0, y=2)

[@milos_radovanovic]

@@sclearDevelopment I get that, but my question wasn't about the difference between minimum and maximum. I wasn't clear, but my question was specifically about the proof that equality was sufficient for his conclusion. Following the same arguments he used a "modified inequality" like 2*(a^2+b^2)(c^2+d^2) >= (ac+bd)^2 all the arguments he made would still hold, but the conclusion would be wrong in my opinion. When we use a greater equal sign all we are saying is precisely what the symbol itself says. The statement that "the fact that equality holds carries a special meaning" must be proven separately from the inequality theorem itself. I'm not advocating for the insistence on great rigour in every video, but conflating inequality and equality (not in the sense of operators but theorems) really confuses things.

[@milos_radovanovic]

I miss Dr Peyam's videos and his bprp colabs. Does anyone know what's he up to?

[@costakeith9048]

@@milos_radovanovic There are proofs of the Cauchy-Schwarz inequality on the Wikipedia page.

[@Meghana_Nallamilli]

@@milos_radovanovicdo you mean that equality may not be attained in all cases and only seems to hold here? By the second inequality you’ve given, √x +√y seems to be bounded above by 2√3 but the actual minimum is √2, so it’s just a loose lower bound

[@sclearDevelopment]

Thanks! I was struggling to understant it for my self-studies because i thought i had to first learn about the inner product space

[@howeworth]

7:38 I would write the equality condition as ad = bc, since c and d can also be 0.

[@teinff7696]

Here's a fun question: What's the maximum number of equidistant points you can place in each dimension and what shape would they take? (example 1D=2 2D=3 3D=4 etc.)

[@paulveba6225]

Please do 100 differential equations kind sir!🙏 You’re the greatest math professor.

[@squashedoranges7949]

Although I understand the point of the video was to use a particular method - I found the solution using substitution (find y in terms of x, then solve for d/dx(√x + √y) = 0 was quite elegant for this problem.

[@h.m.6228]

Is that Putin in the thumbnail?

[@Nothingx303]

No 😂😂😂😂😂😂😂😂😂

[@IoT_]

Да!

[@subratpradhan3581]

😂😂

[@MannKumarGupta]

Yo i just noticed that 😂

[@Happy_Abe]

It’s Cauchy

[@arradrahman8383]

Loved it! ❤

[@SpeedMinxer]

Arrad my pookie ❤

[@sclearDevelopment]

​@@SpeedMinxeridk why I read this with a kratos voice

[@kriegsmesser4567]

Another elementary way of solving it (without lag. mult.) would be: substitute x = (2a)^2, y=(sqrt(2)b)^2. The equations then become: 1) a^2+b^2=1 (just a unit circle) 2) b = -sqrt(2)a + C We must find the maximum C such that both equations are satisfied. The maximum value of C will of course be the value where 2) is tangential to 1). That's equivalent to finding the point where 1) and a line b = 1/sqrt(2)a intersect (this line is perpendicular to 2) and passes through the origin). Plugging in, we get a^2 = 2/3 and b^2 = 1/3. Returning to the original variables, we get x = 8/3 and y = 2/3.

[@AndreTewen]

Other method : x+2y=4 y=(4-x)/2 set f(x)=sqrt(x)+sqrt(y) =sqrt(x)+sqrt((4-x)/2) (domain = ]0;4]) By studying this function we find the maximum is reached for x=8,/3 and the max=sqrt(6).

[@davidseed2939]

but just because the inequality holds, why does that make it a maximum rather than an upper bound

[@blackpenredpen]

Because I also showed that it can achieve that value. X=8/3 and y=2/3

[@Ninja20704]

Equality is achieved if and only if there exists a real constant k such that a=kc and b=kd. Or in another words a/c=b/d. 7:26 In this case, the equality is achievable, so sqrt(6) is the true maximum

[@dededededededewedq]

You show there's an upper bound -> you test that upper bound -> it works -> you proved that's the maximum

[@Onkoe]

interesting! i remember learning something like this at my (relatively rural) Oklahoman high school… i didn’t get it at the time, but you explained it very well! 😄

[@user.glitch]

6:46 can someone explain this part? how is it always >=0?

[@Opaaaa123]

Output of sqrt is defined to be non-negative, meaning its always >=0

[@user.glitch]

@@Opaaaa123 so complex numbers are an exception here?

[@ZipplyZane]

@@user.glitch x and y can't both be negative given the first equation. If one of them is negative, then sqrt(x) + sqrt(y) is a complex number with both real and imaginary components. You cannot compare such numbers with greater than or less than. Therefore a maximum in that case cannot exist. For example, there is no consistent way to say whether 1+2i is greater than or less than 2+i. Neither one is the maximum.

[@Sprezzatura13]

There is a flaw in the solution. The only thing Cauchy-Schartz can say about the values of sqrt(x) + sqrt(y) is that it is never greater than sqrt(6). It doesn’t mean it equals or approaches sqrt(6)

[@BayAdmiral]

Very good video and explanations

[@zactastic4k955]

I don’t know if you take requests but can you do a video on hyper geometric functions I’ve been struggling learning them and you have helped me learn many things so I thought I would ask

[@LITHICKROSHANMS-gw2lx]

please make solution for this one lim(ⁿx) (x→0) ⁿx→infinite tower of 'x'

[@SadregLT]

DNE because we don't know if the number of x are even or odd. We know that the limit of x^x when x approaches 0 equals to 1. In this case we would naturally think that this equals to 1 cuz going from top to bottom we would (I would assume) get 0^0=1 (approaches 1), but at the same time from top to bottom we could get (the value of calculated n powers) 0^1 and then it would equal to 0. So in conclusion lim(ⁿx) (x→0) DNE cuz at the same time it approaches 0 and at the same time it approaches 1 because we can't determine if infinity is odd or even. I hope I answered this for you (My English is not good, so sorry for that)

[@LITHICKROSHANMS-gw2lx]

Thanks

[@SadregLT]

@@LITHICKROSHANMS-gw2lx I saw some mistakes in my answer - reread it if you can.

[@LITHICKROSHANMS-gw2lx]

My answer is zero Because lim(x→0)(ⁿx) SOLUTION:- y=ⁿx....(1) y=((x)^(x)^(x)^(....)))) y=((x)^(ⁿx)) y=((x)^(y)) Taking natural logarithmic function on both sides ln(y)=y(ln(x)) ((1/y))(ln(y))=ln(x) Multiply the negative sign on both sides -(((1/y))(ln(y)))=-ln(x) [Since,→-ln(a)=ln(1/a)] ((1/y)(ln(1/y))=ln(1/x) Modifying the expression in terms of Lambert 'W' function structure in the left hand side (((e)^(ln(1/y)))(ln(1/y)))=(ln(1/x)) Taking Lambert 'W' function on both sides W(((e)^(ln(1/y)))(ln(1/y)))=W(ln(1/x)) ln(1/y)=W(ln(1/x)) Raised the exponent power on both sides 1/y=((e)^(W(ln(1/x)))) Taking the inverse power on both sides So, y=(((e)^(W(ln(1/x)))))^(-1) y=((1)/((e)^(W(ln(1/x)))))....(2) Apply this equation (2) in (1) ⁿx=((1)/((e)^(W(ln(1/x))))) So,the limits become lim(x→0)(((1)/((e)^(W(ln(1/x))))))=0

[@wayneosaur]

But this expression can never reach sqrt(6) can it? -- I get a smaller upper bound of (sqrt(6) + sqrt(3))/2.

[@TundeEszlari]

Good content.

[@lanjiaojiaozhu2745]

Could you do a lecture on Riemann hypothesis, Landau-Siegel zeroes conjecture and also Yitang Zhang's approach. Thanks.

[@rashishsaini50]

???????????????

[@zm4335]

Hi:) Could you elaborate further on the history of the math?

[@MadaraUchihaSecondRikudo]

That sqrt(x)+sqrt(y)

[@coen_xD]

(i) x + 2y = 4. y = 2-x/2 (ii) max {√x + √y} = C. y = (C-√x)^2. where C is a Real number. y = y 2-x/2 =C^2 - 2C√x +x take derivative of both sides -1/2 = -C/√x + 1 √x = 2C/3 x = (4 * C^2) / 9 insert √x into (ii). max { 2C/3 + √y } = C max{ √y } = C/3 √y = C/3 y = (C^2) / 9 insert our found x and y into (i) (4 * C^2)/9 + 2*(C^2)/9 = 4 C^2 = 6 C = ±√6 √x +√y has to be positive, if we work with real numbers. so C =√6

[@Mhmdddm]

Could you please provide some excellent and high-level books on analysis and algebra?

[@andirijal9033]

four metod : Deferential metod, Multiplayer Lagrang metod, Chaucy-Schwarz metod, and AM-GM metod.

[@Subpurgamer]

Why is the life of Boris merch store linking here 😂

[@eeeby_deeby]

Boris himself has just stated on his subreddit that it's been fixed now.

[@martineyles]

Which is older - CS inequality or Calculus? How did they come up with CS?

[@bjornfeuerbacher5514]

Calculus is much older, about 150 years.

[@nmaedu.100]

great teaching technique ✅

[@GreenMeansGOF]

Please do a part 2🙂

[@sepdronseptadron]

is the equality always reachable or are there some scenarios where the system of equations at the end have only complex or no solutions at all? like is the blue work necessary or is a solution also guaranteed by the theorem?

[@sclearDevelopment]

It's always reachable given a/c=b/d. I mean, if there was a third clause saying x

[@actualRocketScientist]

I'm about to apply to Berkeley just so you can be my advisor lol

[@Tzizenorec]

Does this actually prove that a sqrt(x)+sqrt(y)=sqrt(6) solution exists? It looks to me like it only proves that the maximum value can't be greater than sqrt(6), not that it can't be less.

[@ZipplyZane]

That's why he tests that value at the end to get the x and y values. He wasn't quite clear with why he was doing that.

[@RichKeefe-w6c]

Does the question say x and y need to be positive, what happens if one is negative and what is the maximum with [i/j]? Eg x=1,000,000 y=-499,998

[@sclearDevelopment]

Yes they need to be positive in this application of the theorem

[@gdtargetvn2418]

x and y HAS to be positive even if unmentioned or else it will make sqrt(x) + sqrt(y) undefined

[@ayasegoingmeow]

@@gdtargetvn2418 actually, it would just become complex numbers, but then it wouldn't have a comparable value

[@kianushmaleki]

Beautiful

[@franepoljak9605]

is it a coincidence that you got x + 2y = 4 when calculating x and y that gives maximum value (√6), same as the first equation?

[@kolirahman8948]

a^x=a^y+2^z here(a,x,z,y are positive intigre.) now find the sum of all solution (ayz/(x-1)^2). Please solve this sir..........

[@geekonomist]

That inequality simply says that the sum cannot be more than some value. That does not mean that it is even possible for the sum to reach that value. It might be that the equation minimum can only be below this value.

[@josenobi3022]

This. I don’t know why this isn’t the top comment

[@blackpenredpen]

7:26

[@josenobi3022]

@@blackpenredpen that should have been done before the other steps. You also said the minimum of x²+y² was 16/5 with an incorrect reasoning

[@geekonomist]

@@blackpenredpen ok. I did not oook that far. Thank you for the correction

[@epsilia3611]

You claim that this inequality gives us a maximum value for the function f(x)=sqrt(x)+sqrt(2-x/2). But it is a maximum if it is reached by the function at the first place. You ask yourself the question "but when is sqrt(6) obtained though ?" after wondering if what you are calculating is reached beforehand. WARNING : IT DOESN'T WORK FOR ALL SORTS OF INEQUALITIES. Explanation : It turns out that the Cauchy-Schwarz inequality is very useful mostly because there is an equality case ! If there were not an equality case, the max wouldn't be even shown in the inequality. So I think what's necessary, and so important to explain here is how the equality case does have its role into knowing that the C-S inequality IMPLIES that the majorant extracted out of it WILL be a max. Conter-example : we can say that f(x) < or = 3, but since we don't know if that equality could hold for a certain x, it doesn't imply that we get a certain x such that f(x)=3 !

[@as4study]

From India🇮🇳

[@SuperDeadparrot]

I would’ve just used Lagrange multipliers.

[@pashaw8380]

It’s become almost as instinctive and intuitive as the EVT at this point. Finding the critical point of a function is straightforward-just take its derivative and set it to zero. While Cauchy and Hölder’s inequalities are invaluable for proving and developing theorems, they don’t offer the same intuitive grasp. These inequalities are potent tools, extending far beyond locating the extrema; they can even meticulously unravel systems of nonlinear equations. However, I wouldn't use them for pinpointing the extrema of a function.

[@jyotsanabenpanchal7271]

Please make a video on stolz cesaro theorem please 🥺🥺🥺❤️

[@erwanquintin3057]

MRS=p1/p2, ECON 101

[@rogerphelps9939]

Substitute for y in the frst equattion and tthen differentiate setting the result to zero.

[@gaetanramos7903]

Consider this stolen on behalf of "Omg this is so much better than calculus".

[@cdkw2]

nice!

[@gregevgeni1864]

Nice

[@KAS_ACADEMY]

Math is illegal İn the USA

[@trongquang3128]

nice

[@GeoPeron]

ahhh I used calculus, I'm sorry 😭

[@Glorious_Potato]

How do we know sqrt(6) is the maximum and not just an upper bound of sqrt(x) + sqrt(y)?

[@jarikosonen4079]

Is there any case for Cauchy-Schwarz that the equality-case is not satisfied?

[@Danmreke]

Can you please explain this equation? x^4 + 12x^3 + 42x^2 + 36x + 25

[@chancia8990]

lol it's easier just doing the calc 3

[@Oskar-zt9dc]

the "cauchy-swanz-ineuality" 😂

[@Selma_Benacer]

From Algeria 🇩🇿 Thank u teacher 🤗🤗🤗

[@broytingaravsol]

easy

[@fadydavis7457]

That mathematician in the thumbnail was the greatest enemy of US

[@chrisglosser7318]

Solved using polar coordinates

[@chrisglosser7318]

Doesn’t have the ambiguity of Cauchy-Schwartz

[@MikehMike01]

this hardly seems useful

[@bsharp111]

arent you just the most enjoyable person

[@sajuvasu]

First pls pin..❤

[@MrSomethingdark]

dude still hasn't learned how to talk properly

[@Lolwutdesu9000]

I fail to see how root (x) + root (y) must always be greater than or equal to zero. The roots of x and y can be both positive and negative, and the sum of two negative numbers (or even one positive or negative) can still be negative. That was a very hand-wavey reasoning IMO.

[@howeworth]

The output of the square root function is nonnegative by definition.

[@samuelagaga4348]

To maximize xyx^yxy given the constraint x+y=8x + y = 8x+y=8, we can use calculus and the method of Lagrange multipliers. Step 1: Express the Problem We want to maximize the function: f(x,y)=xyf(x, y) = x^yf(x,y)=xy subject to the constraint: g(x,y)=x+y−8=0g(x, y) = x + y - 8 = 0g(x,y)=x+y−8=0 Step 2: Use the Method of Lagrange Multipliers The method of Lagrange multipliers tells us that at the maximum point, the gradients of f(x,y)f(x, y)f(x,y) and g(x,y)g(x, y)g(x,y) are proportional. Therefore, we set up the following system of equations: ∇f=λ∇g abla f = \lambda abla g∇f=λ∇g where λ\lambdaλ is the Lagrange multiplier. Compute the gradients: ∇f(x,y)=(∂f∂x,∂f∂y)=(yxy−1,xyln⁡(x)) abla f(x, y) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} ight) = \left( yx^{y-1}, x^y \ln(x) ight)∇f(x,y)=(∂x∂f​,∂y∂f​)=(yxy−1,xyln(x)) ∇g(x,y)=(∂g∂x,∂g∂y)=(1,1) abla g(x, y) = \left( \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} ight) = (1, 1)∇g(x,y)=(∂x∂g​,∂y∂g​)=(1,1) Thus, we have: (yxy−1,xyln⁡(x))=λ(1,1)\left( yx^{y-1}, x^y \ln(x) ight) = \lambda (1, 1)(yxy−1,xyln(x))=λ(1,1) This gives us two equations: yxy−1=λyx^{y-1} = \lambdayxy−1=λ xyln⁡(x)=λx^y \ln(x) = \lambdaxyln(x)=λ Step 3: Solve the System of Equations From the two equations: yxy−1=xyln⁡(x)yx^{y-1} = x^y \ln(x)yxy−1=xyln(x) Divide both sides by xy−1x^{y-1}xy−1: y=xln⁡(x)y = x \ln(x)y=xln(x) Step 4: Substitute into the Constraint We know y=8−xy = 8 - xy=8−x from the constraint x+y=8x + y = 8x+y=8. Therefore: 8−x=xln⁡(x)8 - x = x \ln(x)8−x=xln(x) This is a transcendental equation that we need to solve for xxx. Step 5: Solve 8−x=xln⁡(x)8 - x = x \ln(x)8−x=xln(x) This equation cannot be solved analytically in a simple way, so we solve it numerically or approximate it by inspection. To find the value of xxx that satisfies the equation, we can try some values: If x=2x = 2x=2, 8−2=68 - 2 = 68−2=6 and 2ln⁡(2)≈1.3862 \ln(2) \approx 1.3862ln(2)≈1.386. If x=3x = 3x=3, 8−3=58 - 3 = 58−3=5 and 3ln⁡(3)≈3.2953 \ln(3) \approx 3.2953ln(3)≈3.295. If x=4x = 4x=4, 8−4=48 - 4 = 48−4=4 and 4ln⁡(4)≈5.5454 \ln(4) \approx 5.5454ln(4)≈5.545. If x=5x = 5x=5, 8−5=38 - 5 = 38−5=3 and 5ln⁡(5)≈8.0465 \ln(5) \approx 8.0465ln(5)≈8.046. x=3x = 3x=3 seems to work best because 3⋅ln⁡(3)3 \cdot \ln(3)3⋅ln(3) is close to 555. Step 6: Calculate xyx^yxy at x=3x = 3x=3 If x=3x = 3x=3, then y=5y = 5y=5. So: xy=35=243x^y = 3^5 = 243xy=35=243 Conclusion The maximum value of xyx^yxy given x+y=8x + y = 8x+y=8 is 243\boxed{243}243​ when x=3x = 3x=3 and y=5y = 5y=5.