x + 2y = 4 2y = 4 - x y = 2 - .5x let A = sqrt(x) + sqrt(y) A = sqrt(x) + sqrt(2 - .5x) A' = 1/[2sqrt(x)]- 1/[4sqrt(2 - .5x)] let A' = 0 1/[2sqrt(x)] = 1/[4sqrt(2-.5x)] 2sqrt(x) = 4sqrt(2 - .5x) 4x = 16(2 - .5x) x = 8 - 2x 3x = 8 x = 8/3 when x = 8/3 a maximum occurs We need to check endpoints as well: C1: x = 0 A = sqrt(0) + sqrt(2) A = sqrt(2) ~= 1.414 C2: x = 8/3 A = sqrt(8/3) + sqrt(2 - 4/3) A = 2sqrt(2/3) + sqrt(2/3) A = 3sqrt(2/3) = sqrt(6) ~= 2.45 C3: y = 0, i.e. x = 4 A = sqrt(4) + sqrt(2 - 2) A = 2 So the maximum of sqrt(x) + sqrt(y) occurs when x = 8/3 and is equal to sqrt(6).
Sir please make a video on use of and , or in mathematics .means its meaning like in set theory or means union and and meaning intersection.in probability different so please make a video on that.
Very clever! I did it a different way, bit long-winded but it worked. A = x^3 (3 - x) -A = x^4 - 3x^3 Suppose the quartic on the right touches a line y = -b at x = a. (x-a) will be a repeated factor. Write x^4 - 3x^3 + b = (x - a)^2 (x^2 + cx + d) [edited, see below] = (x^2 - 2ax + a^2)(x^2 + cx + d) = x^4 + (-2a + c) x^3 + (d - 2ac + a^2) x^2 + (-2ad + a^2c) x + a^2 d Equating x^3 coefficients c - 2a = -3 c = 2a - 3 Equating x coeffs a^2c - 2ad = 0 ac = 2d d = ac/2 d = a(2a - 3)/2 Substitute back in x^2 coeff a(2a-3)/2 - 2a(2a-3) + a^2 = 0 (2a-3)/2 - 2(2a-3) + a = 0 a - 3/2 - 4a + 6 + a = 0 2a = 9/2 a = 9/4, the solution for x Equating constant coeffs b = area = a^2 a(2a-3)/2 = (729/64)(3/2)/2 = 2187 / 256
@@pwmiles56 I like your method, although I believe you forgot to shift the function by -b before equating it, since when you equated the area function and the factored form on the right, the equality doesn't hold as can be seen by substituting x = a. Feel free to correct me if I am wrong, good work nonetheless!
@islam2681 You are quite right. The equation should read x^4 - 3x^3 + b = (x - a)^2 (x^2 + cx + d) (b will be A, of course, but I am leaving it free for now) Thanks!
Nice video, and thank you for being careful with the details (e.g., you verified the variables used in the application of AM-GM were non-negative, you made sure the upper bound was actually a maximum, and you made sure that in the claimed equality case, the x was in the required interval).
The academic year has just ended in NZ and highschool students and their parents are up in arms about an optimisation problem involving the min surface area for a cylinder of given volume when the students had not been taught how to differentiate f(r)=1/r. But the answer is arguably the most famous cylinder shape in the history of Mathematics, and like so many of these problems, no calculus is required.
How the hell do you not teach students the most basic derivative of x^n??? we literally learned that in the first week we were first introduced to the derivative (wow, it works for integers. wow it works for rationals because chain rule. wow, it works for real numbers because density of rationals and since if qn converges to r, on any bounded interval, x^qn converges to x^r uniformly and hence d/dx x^qn converges to d/dx x^r. Then just expand the interval to include an arbitrary point and you're done). Also, you can *easily* do 1.x directly from first principles.
i don't know if you guys will believe me or not, but i thought about am gm just by seeing the thumbnail ( that may be becuz I've done questions like that before ) and definitely it's a very clean way to avoid calculus, thanks for spreading such solutions
So much easier to use calculus. A = x³(3-x) = -x⁴ + 3x³. Differentiating we get A' = -4x³ + 9x². Setting that = 0, we have 4x³ = 9x², or 4x = 9. So, x = 9/4. The area would then be (3 - 9/4)(9/4)³. That comes to 2187 / 256, the same answer BPRP got.
well the point here is to explore and learn other possible ways to solve the problem instead of just looking for the easiest way. We never know if there may be a similar problem where using calculus may be much harder than this method, such as optimisation problems with 3 or more variables
@@Ninja20704 I know, I was just making a point. So often people think calculus is hard, and avoiding it makes it easier. This is a good counter-example.
Задача очень простая. S=(3-x)x^3 Находим производную площади. S’=9x^2-4x^3 Находим экстремум. x=9/4 Проверяем знак производной справа и слева от точки. При данном значение Х площадь имеет наибольшее значение.
I agree. You can't appeal to AM GM and then say you didn't use calculus. Proving GM is always less than AM requires a certain amount of calc and/or real analysis.
You want the maximum to be a constant, so there should be an idea of cancelling all x terms, and converting from product to sum through AM-GM is a logical way.
The AM-GM inequality is very often useful for solving maximization problems - because it is an inequality and hence tells you how big something can get maximally.
@Bof-q6l Yes, I let the x being x and changed -x to -3x (also 3 to 9) with the benefit of not having fractions, just natural numbers! "Easier", at least for me. With the same result of 9/4.
The argumentation why you can just arbitrarily cut 3 in 3 thirds is beyond me. Maybe I'm missing something. I see why it works, as you got the same weight on (3 - x) and x³ as the two sides of the rectangle. But what if you have (3 - 2x) ? Then you have 2/3 x each. Similar result: 9/8. LHS = (3 - 2x) (2/3 x)³ = 8/27 (3 - 2x) x³ / RHS = 27/8 (3/4)^4 = 3^7 / 2^11 Scaling it to the original area formula: (3 - 2x) x³
The AM GM inequality still holds and it simplifies to an expression that matches the desired expression, and he verified that the resulting constant is a possible value of the expression.
@@civilizationkills3138 I wanted to edit the former post, as I get it now. This method seeked a solution for the 4 dimensional cube with sizes (3 - x), x/3, x/3, x/3, and all these values are 3/4.
Is there a general way of knowing when a problem of this type is solvable using this method? Could you do this for any (positive) polynomial on any interval?
Sorry that is too complicated, requires knowledge of the AM-GM and the very smart trick of replacing the x by x/3, which is based on hindsight that the AMGM rule can be used. The simple and straightforward solution is using the derivative to find maximum and minimum values. dA(x)/dx =A‘(x)= 9x^2 - 4x^3 => A‘(x)= x^2*(9-4x) 1) solving for A‘(x)=0 to find maximum/minimum of A: A‘(x)= 0 => x=0 or x=9/4 2) check whether these values correspond to minimum or maximum: For x=0 => A(0)=0 : obviously a minimum For x=9/4: The term x^2 of A’(x) is always positive. The term (9-4x) of A’(x) changes sign from positive to negative by passage by x=9/4 => A(x) is increasing for 0 ≤ x ≤ 9/4, and decreasing for 9/4 ≤ x ≤ 3 => 9/4 defines a maximum of A(x), and Max[A(x)] = A(9/4)=2187/256
@fridolfwalter2256 I thought as (3x-x²)(x²) is the same as the initial one, the max of this will give the answer The max value happens when both terms are equal(as he said in video) x =0 is neglected as that is the minimum Which gives x =1.5 x is between 0 and 3 and thus is also an achievable value giving area 81/16 which is incorrect
I think in this case the max of (3x - x^2)(x^2) isn't necessarily when 3x - x^2 = x^2 because of the presence of the x^2. It is true that for the product of two linear factors like x(3-x) the maximum is when x = 3-x but this doesn't necessarily hold for more complicated expressions.
@cyberduck027 I got it. Its because the sum must be a constant. In your example 3-x+x=3 and that's why the equality holds. Same for bprp's expression. Thank you! I wouldn't have been able to realize it without your comment
Check the Calculus way!? xxx(3-x) = 3x^3 - x^4 , d/dx that function = 0 for maximu or minima. 9X^2 - 4x^3 = 0 and doing second derivative test of that to say maxma or minima 18x - 12x^2 or 18 - 12x being positive at 9x^2 - 4x^3 or at 9 - 4x point or x = 9/4 point. If you know Calculas don't be afraid to use it to quickly find x= 9/4 location and a maxima at that point from Calculus 2nd derivative. Yes, general harder way mathematics without Calculus is neat but Calculus knoledge is faster and neater.
Hello sir, please show how to solve the following integral: integral from x to y of sin(xt)/cos(yt) dt I have tried substitution method and integration by parts and neither worked. Thank you! And keep up the good work :)
He wasn't trying to preserve the inequality. He was trying to pick values for b, c and d to make the problem easier to solve. He could have easily chosen b=1, c=2 and d=3 and the inequality would still hold--it just wouldn't help solve the problem, though.
The inequality holds for any a, b, c, d. It holds for a = 3-x, b = x, c = x, d = x, but it also holds for a = 3-x, b = x/3, c = x/3, d = x/3. You only have to insert the same values on _both_ sides of the inequality.
At the end you had (x-3) = x/3 because a=b=c=d. You didn't even need to go through the 4th root((x-3)x^3/27 part. And also, why is the "=" true when a=b=c=d?
No, it's a = 3-x, b = x/3, c = x/3, d = x/3, starting from 6:15 and going on from there. Using a=3-x, b=x, c=x, d=x does not help in solving this problem, as he explains in the video.
Taking f(x)=x³ we have f'(x)=3x² and thus f(a)=3a². Now can calculate the tangent line at the point a which is y=3a²x-2a³, and we see that this line intercepts the x axis at the point x=⅔a. Now we can finally calculate the area of the triangle under the line (a-⅔a)a³/2 since the base is the segment that goes from ⅔a to a and the height is a³. In the end we say that the area is a⁴/6 which is equal to 1.5 or 3/2 when a=9^(1/4). You could have also used the integral from ⅔a to a of 3a²x-3a³ with respect of x and then forcing the result of the integral to be equal to 3/2
I was thinking that inequality wasn't quite right. The area can't be 0 so x cannot be 0 or 3, so the restriction on x is 0 < x < 3. The answer at the end was still correct, though, so maybe that didn't matter in this instance.
@@LordQuixote Yes and we don't want to include those extrema, so we restrict our domain for x to the open interval (0, 3) instead of a closed interval [0, 3].
@@LordQuixote The problem specifies a rectangle fit under a curve. At x = 0 and x = 3, we don't have a rectangle. So it's an important restriction to have.
@@jmwild1 Yeah, no, not really. In a general problem, if the best answer is zero, you'll lose that if you restrict the domain and get an incorrect answer. (For example, what's the most optimal area to not lose money? Ans: 0)
I get that this is a stipulation for the problem, but I don't understand when or why you would solve a problem like this and not be willing or able to use calculus.
It's obvious when or why. It is to demonstrate how such a problem can be solved using a different method - one that many would never have thought of, and/or may not be aware of.
@@donmoore7785 My first university math book "Calculus: A Complete Course" had a section very early in the book where it showed how the area of a circle could be derived by thinking of it as adding together lots of isosceles triangles, and then taking the limit of the number of triangles. That method was very insightful.
@ jamescollier3 @ donmoore7785 @ Peter_1986 -- None of you got the point of Skyler827 that why would it occur to a solver to use this method *without it being brought up.*
Don't learn Calculus and abandon it! Mechanical or Dimensions Architects professional people don't go back to basic mathematics to solve their Physics problems. Otherwise not using Calculus in the Canon Ball trajectory problems with initial velocities just became difficult of Calculus never existed.
Could you do a video on the differentiation of f(x) =|x|? I can't find an explanation for why it's x /|x|. (it makes sense but I'd like to know how you get there)
He does have it in some of his videos but I’m not sure which ones. But the simplest way is rewrite |x| as sqrt(x^2) then differentiate the usual way, then replace the sqrt(x^2) back with |x| in the final answer. So d/dx[|x|] =d/dx[sqrt(x^2)] =1/[2sqrt(x^2)] * 2x =x/sqrt(x^2) =x/|x|
The definition of absolute value is : |x|=sqrt(x^2) If you want to differentiate |x| you can just take the derivative of sqrt(x^2) which is x/sqrt(x^2) = x/|x|
Huh? No, he did not divide the equation by 3 at 5:32, what are you talking about? And there is no equality at 5:32, only an inequality. Are we talking about the same video???
He was just trying to figure out what to choose for a, b, c, and d. He was looking for a way to remove the x from the right side by choosing values for b, c, and d that would eliminate the -x from the (3 - x) he had for a. Because he chose those for b, c, and d, he then had to do that to both sides of his inequality. This way, he had a constant value on one side and then was able to proceed until he got his original equation on one side
@@lawrencejelsma8118well the point here is to explore and learn other possible ways to solve the problem instead of just looking for the easiest way. We never know if there may be a similar problem where using calculus may be much harder than this method, such as optimisation problems with 3 or more variables
Doing it the calculus way shows that the maximum area (where the derivative = 0) occurs at x=3/4 (0.75) and not at x=9/4. I guess that the substitution should be reversed at the very end by dividing by 3.
"Doing it the calculus way shows that the maximum area (where the derivative = 0) occurs at x=3/4 (0.75) and not at x=9/4." Actually, doing it the calculus way also gives x = 9/4. I don't know how you got x = 3/4. Could you please show your calculations? "I guess that the substitution should be reversed at the very end by dividing by 3." No, that makes no sense, why should one do that? Which "substitution" are you even talking about?
@FaranAiki consider four positive quantities (3 - x)/1 , x/3 , x/3 , x/3 with sum = 3 when we apply AM - GM inequality on these quantities we get the result. in AM - GM inequality equality holds iff all quantities are equal hence (3 - x)/1 = x/3 in other words ratio between (3 - x) and x is 1:3 and since their sum is 3 we can say that we have to divide 3 in the ratio 1:3 other examples - (1) find the maximum value of (x^3){ (20 - x)^2 } if 0 < x < 20 we have to divide 20 in the ratio 3 :2 x = 12 , 20 - x = 8 maximum value is (12^3)(8^2) = 110592 (2) find the maximum value of (x^2)(y^3) if 3 x+2 y = 1 , x > 0 , y > 0 we have to divide 1 in the ratio 2 : 3 3 x = 2/5 , 2 y = 3/5 or x = 2/(3)(5) , y = 3/(2)(5) maximum value is 3/6250 (3) find the maximum value of xy if x+y = 8 , x > 0 , y > 0 we have to divide 8 in the ratio 1:1 x = 4 , y = 4 maximum value is 16
If x+2y=4, then find the max of sqrt(x)+sqrt(y): th-cam.com/video/YSJzpQQSOZw/w-d-xo.html
God job bro
x + 2y = 4
2y = 4 - x
y = 2 - .5x
let A = sqrt(x) + sqrt(y)
A = sqrt(x) + sqrt(2 - .5x)
A' = 1/[2sqrt(x)]- 1/[4sqrt(2 - .5x)]
let A' = 0
1/[2sqrt(x)] = 1/[4sqrt(2-.5x)]
2sqrt(x) = 4sqrt(2 - .5x)
4x = 16(2 - .5x)
x = 8 - 2x
3x = 8
x = 8/3
when x = 8/3 a maximum occurs
We need to check endpoints as well:
C1: x = 0
A = sqrt(0) + sqrt(2)
A = sqrt(2) ~= 1.414
C2: x = 8/3
A = sqrt(8/3) + sqrt(2 - 4/3)
A = 2sqrt(2/3) + sqrt(2/3)
A = 3sqrt(2/3) = sqrt(6) ~= 2.45
C3: y = 0, i.e. x = 4
A = sqrt(4) + sqrt(2 - 2)
A = 2
So the maximum of sqrt(x) + sqrt(y) occurs when x = 8/3 and is equal to sqrt(6).
2.25*2.25*2.25*.75=8. ... ?
Sir please make a video on use of and , or in mathematics .means its meaning like in set theory or means union and and meaning intersection.in probability different so please make a video on that.
I think he's good at Math guys
That is not a sentence. For one thing, you are addressing "guys." Try the next sentence. I think he's good at math, guys.
@@forcelifeforce Sadly, nobody cares.
Nah I disagree
@@forcelifeforce top 10 'we dont give a fuck' moments
@@generalkenobi9803 I alongside many people do care, you don't know how infuriating it is to recognise improper grammar and sentence structure.
Very clever! I did it a different way, bit long-winded but it worked.
A = x^3 (3 - x)
-A = x^4 - 3x^3
Suppose the quartic on the right touches a line y = -b at x = a. (x-a) will be a repeated factor. Write
x^4 - 3x^3 + b = (x - a)^2 (x^2 + cx + d) [edited, see below]
= (x^2 - 2ax + a^2)(x^2 + cx + d)
= x^4
+ (-2a + c) x^3
+ (d - 2ac + a^2) x^2
+ (-2ad + a^2c) x
+ a^2 d
Equating x^3 coefficients
c - 2a = -3
c = 2a - 3
Equating x coeffs
a^2c - 2ad = 0
ac = 2d
d = ac/2
d = a(2a - 3)/2
Substitute back in x^2 coeff
a(2a-3)/2 - 2a(2a-3) + a^2 = 0
(2a-3)/2 - 2(2a-3) + a = 0
a - 3/2 - 4a + 6 + a = 0
2a = 9/2
a = 9/4, the solution for x
Equating constant coeffs
b = area = a^2 a(2a-3)/2
= (729/64)(3/2)/2 = 2187 / 256
Nice spacing
@@pwmiles56 I like your method, although I believe you forgot to shift the function by -b before equating it, since when you equated the area function and the factored form on the right, the equality doesn't hold as can be seen by substituting x = a. Feel free to correct me if I am wrong, good work nonetheless!
@islam2681 You are quite right. The equation should read
x^4 - 3x^3 + b = (x - a)^2 (x^2 + cx + d)
(b will be A, of course, but I am leaving it free for now)
Thanks!
Great work
Nice video, and thank you for being careful with the details (e.g., you verified the variables used in the application of AM-GM were non-negative, you made sure the upper bound was actually a maximum, and you made sure that in the claimed equality case, the x was in the required interval).
Thank you!
The academic year has just ended in NZ and highschool students and their parents are up in arms about an optimisation problem involving the min surface area for a cylinder of given volume when the students had not been taught how to differentiate f(r)=1/r. But the answer is arguably the most famous cylinder shape in the history of Mathematics, and like so many of these problems, no calculus is required.
How the hell do you not teach students the most basic derivative of x^n??? we literally learned that in the first week we were first introduced to the derivative (wow, it works for integers. wow it works for rationals because chain rule. wow, it works for real numbers because density of rationals and since if qn converges to r, on any bounded interval, x^qn converges to x^r uniformly and hence d/dx x^qn converges to d/dx x^r. Then just expand the interval to include an arbitrary point and you're done). Also, you can *easily* do 1.x directly from first principles.
Nice method but I'd still use calculus though!
Yea, I don't recall being taught this AM-GM method. I'll have to look it up. But I do remember my basic calculus, so derivative for the win!
@@Rev03FFL its op!
Na minha opinião, um dos melhores da Internet: objetivo, claro e resolve problemas valorosos! Obrigado, companheiro!
Thank you. I must admit, doing the calculus is way easier! In any case, the AM-GM is perhaps a better testing one's resilience in thinking forward.
where does am gm come from? never heard of it😢
Solve enough number of problems and u will find out the importance of am gm
Calculus master is back with more awesome stuff 🙌🤗
i don't know if you guys will believe me or not, but i thought about am gm just by seeing the thumbnail ( that may be becuz I've done questions like that before ) and definitely it's a very clean way to avoid calculus, thanks for spreading such solutions
Oh man that is a satisfying method. Bravo.
So much easier to use calculus. A = x³(3-x) = -x⁴ + 3x³. Differentiating we get A' = -4x³ + 9x². Setting that = 0, we have 4x³ = 9x², or 4x = 9. So, x = 9/4.
The area would then be (3 - 9/4)(9/4)³. That comes to 2187 / 256, the same answer BPRP got.
well the point here is to explore and learn other possible ways to solve the problem instead of just looking for the easiest way. We never know if there may be a similar problem where using calculus may be much harder than this method, such as optimisation problems with 3 or more variables
@@Ninja20704 I know, I was just making a point. So often people think calculus is hard, and avoiding it makes it easier. This is a good counter-example.
Задача очень простая.
S=(3-x)x^3
Находим производную площади.
S’=9x^2-4x^3
Находим экстремум.
x=9/4
Проверяем знак производной справа и слева от точки.
При данном значение Х площадь имеет наибольшее значение.
Vravo..I did the same ..
Прочти условие внимательнее. Производные использовать нельзя.
Amax using derivative
A = (3-x)*y
= (3-x)*x^3
= 3x^3 - x^4
dA/dx = 9x^2 - 4x^3 = 0
x^2*(9 - 4x) = 0
=> x = 0 (-> Amin=0)
x = 9/4 (-> Amax)
Amax = (3 - 9/4)*(9/4)^3 = 2187/256
More generally, the largest rectangle bounded by the x-axis, y=x^3, and the vertical line x=a has area (27/256)a^4 and meets the curve at x=(3/4)a
“Without calculus” proceeds to use a result from Real Analysis without justification 😂
wdym. everything made sense to me. all very basic stuff and I'm a sophomore in high school
I agree. You can't appeal to AM GM and then say you didn't use calculus. Proving GM is always less than AM requires a certain amount of calc and/or real analysis.
Beautifully done, what a piece of artistic work.
Thank so much 🎉
i started trembling and sweating at the no calc statement
why would am gm come in your head in the first place????????
You want the maximum to be a constant, so there should be an idea of cancelling all x terms, and converting from product to sum through AM-GM is a logical way.
The AM-GM inequality is very often useful for solving maximization problems - because it is an inequality and hence tells you how big something can get maximally.
Because my viewers told me about this 😆
How elegant!
How is the geometric mean the maximum? Please explain.
Isn't it a bit easier to use (9 - 3x) instead of thrice x/3? => AM = ((9 - 3x) + x + x + x) = 9 / 4.
LHS = (9 - 3x) x³ / RHS = (9/4)^4 => (3 - x) x³ = 1/3 (3^8/2^8) = 3^7/2^8 = 2187 / 256.
Same thing
You just multiplied all by 3
@Bof-q6l Yes, I let the x being x and changed -x to -3x (also 3 to 9) with the benefit of not having fractions, just natural numbers! "Easier", at least for me. With the same result of 9/4.
The argumentation why you can just arbitrarily cut 3 in 3 thirds is beyond me. Maybe I'm missing something.
I see why it works, as you got the same weight on (3 - x) and x³ as the two sides of the rectangle.
But what if you have (3 - 2x) ? Then you have 2/3 x each. Similar result: 9/8.
LHS = (3 - 2x) (2/3 x)³ = 8/27 (3 - 2x) x³ / RHS = 27/8 (3/4)^4 = 3^7 / 2^11
Scaling it to the original area formula: (3 - 2x) x³
The AM GM inequality still holds and it simplifies to an expression that matches the desired expression, and he verified that the resulting constant is a possible value of the expression.
@@civilizationkills3138 I wanted to edit the former post, as I get it now. This method seeked a solution for the 4 dimensional cube with sizes (3 - x), x/3, x/3, x/3, and all these values are 3/4.
Is there a general way of knowing when a problem of this type is solvable using this method? Could you do this for any (positive) polynomial on any interval?
Sorry that is too complicated, requires knowledge of the AM-GM and the very smart trick of replacing the x by x/3, which is based on hindsight that the AMGM rule can be used.
The simple and straightforward solution is using the derivative to find maximum and minimum values.
dA(x)/dx =A‘(x)= 9x^2 - 4x^3
=> A‘(x)= x^2*(9-4x)
1) solving for A‘(x)=0 to find maximum/minimum of A:
A‘(x)= 0 => x=0 or x=9/4
2) check whether these values correspond to minimum or maximum:
For x=0 => A(0)=0 : obviously a minimum
For x=9/4:
The term x^2 of A’(x) is always positive.
The term (9-4x) of A’(x) changes sign from positive to negative by passage by x=9/4
=> A(x) is
increasing for 0 ≤ x ≤ 9/4, and
decreasing for 9/4 ≤ x ≤ 3
=> 9/4 defines a maximum of A(x), and
Max[A(x)] = A(9/4)=2187/256
what if you can rotate the rectangle? (keeping 0 < x = 0)
I did (3-x)x³ = (3x-x²)(x²)
Both must be equal so,
3x-x²=x²
x=1.5
What did I do wrong?
Can you explain your thought process further?
@fridolfwalter2256 I thought as (3x-x²)(x²) is the same as the initial one, the max of this will give the answer
The max value happens when both terms are equal(as he said in video)
x =0 is neglected as that is the minimum
Which gives x =1.5
x is between 0 and 3 and thus is also an achievable value giving area 81/16 which is incorrect
I think in this case the max of (3x - x^2)(x^2) isn't necessarily when 3x - x^2 = x^2 because of the presence of the x^2. It is true that for the product of two linear factors like x(3-x) the maximum is when x = 3-x but this doesn't necessarily hold for more complicated expressions.
@cyberduck027 I got it. Its because the sum must be a constant. In your example 3-x+x=3 and that's why the equality holds. Same for bprp's expression.
Thank you! I wouldn't have been able to realize it without your comment
Check the Calculus way!? xxx(3-x) = 3x^3 - x^4 ,
d/dx that function = 0 for maximu or minima. 9X^2 - 4x^3 = 0 and doing second derivative test of that to say maxma or minima 18x - 12x^2 or 18 - 12x being positive at 9x^2 - 4x^3 or at 9 - 4x point or x = 9/4 point.
If you know Calculas don't be afraid to use it to quickly find x= 9/4 location and a maxima at that point from Calculus 2nd derivative.
Yes, general harder way mathematics without Calculus is neat but Calculus knoledge is faster and neater.
Hello sir, please show how to solve the following integral:
integral from x to y of sin(xt)/cos(yt) dt
I have tried substitution method and integration by parts and neither worked. Thank you! And keep up the good work :)
can someone explain why you can just divide by 3 on a, b, c? I've wouldn't think that preserves the equation or inequality?😮 Thanks
oops: b, c, d
He wasn't trying to preserve the inequality. He was trying to pick values for b, c and d to make the problem easier to solve. He could have easily chosen b=1, c=2 and d=3 and the inequality would still hold--it just wouldn't help solve the problem, though.
The inequality holds for any a, b, c, d. It holds for a = 3-x, b = x, c = x, d = x, but it also holds for a = 3-x, b = x/3, c = x/3, d = x/3. You only have to insert the same values on _both_ sides of the inequality.
Instead of finding the maximum area of the rectangle, he's finding the maximum of 1/27th of the area of the rectangle. Happens at the same place.
He is solving for the particular value of x where (3-x)=x/3 because then 3-x +x/3+x/3+x/3=the constant 3 .
How to check is this is a maximum area?
Huh?! The video showed that this is the maximum area, there is no need to check this again.
But if you insist, one could use calculus.
Very clever!
Learned something new 😊❤
splendid ❤
I got as far as an inequality was going to be the approach.
At the end you had (x-3) = x/3 because a=b=c=d.
You didn't even need to go through the 4th root((x-3)x^3/27 part.
And also, why is the "=" true when a=b=c=d?
because the quartic root of a^4 is the same as 4a/4
8:33 shouldnt it be x=3-x instead of 3-x=x/3 ??? it's a=b=c=d and a=3-x, b=x, c=x, d=x
No, it's a = 3-x, b = x/3, c = x/3, d = x/3, starting from 6:15 and going on from there. Using a=3-x, b=x, c=x, d=x does not help in solving this problem, as he explains in the video.
Please solve: Form a tangent line to the curve x^3 at point “a” such that the area under the line from its x intercept to “a” is 1.5 areaunits
Taking f(x)=x³ we have f'(x)=3x² and thus f(a)=3a².
Now can calculate the tangent line at the point a which is y=3a²x-2a³, and we see that this line intercepts the x axis at the point x=⅔a.
Now we can finally calculate the area of the triangle under the line (a-⅔a)a³/2 since the base is the segment that goes from ⅔a to a and the height is a³.
In the end we say that the area is a⁴/6 which is equal to 1.5 or 3/2 when a=9^(1/4).
You could have also used the integral from ⅔a to a of 3a²x-3a³ with respect of x and then forcing the result of the integral to be equal to 3/2
thats awesome
yes
I was thinking that inequality wasn't quite right. The area can't be 0 so x cannot be 0 or 3, so the restriction on x is 0 < x < 3. The answer at the end was still correct, though, so maybe that didn't matter in this instance.
3*0^3=0 or 0*3^3=0: the two extreme areas (basically a horizontal line or a vertical line)
@@LordQuixote Yes and we don't want to include those extrema, so we restrict our domain for x to the open interval (0, 3) instead of a closed interval [0, 3].
@@jmwild1 Why? It's an unnecessary restriction.
@@LordQuixote The problem specifies a rectangle fit under a curve. At x = 0 and x = 3, we don't have a rectangle. So it's an important restriction to have.
@@jmwild1 Yeah, no, not really. In a general problem, if the best answer is zero, you'll lose that if you restrict the domain and get an incorrect answer. (For example, what's the most optimal area to not lose money? Ans: 0)
I get that this is a stipulation for the problem, but I don't understand when or why you would solve a problem like this and not be willing or able to use calculus.
to learn different things
It's obvious when or why. It is to demonstrate how such a problem can be solved using a different method - one that many would never have thought of, and/or may not be aware of.
@@donmoore7785 My first university math book "Calculus: A Complete Course" had a section very early in the book where it showed how the area of a circle could be derived by thinking of it as adding together lots of isosceles triangles, and then taking the limit of the number of triangles. That method was very insightful.
@ jamescollier3
@ donmoore7785
@ Peter_1986 -- None of you got the point of Skyler827 that why would it occur to a solver to use this method *without it being brought up.*
Don't learn Calculus and abandon it! Mechanical or Dimensions Architects professional people don't go back to basic mathematics to solve their Physics problems. Otherwise not using Calculus in the Canon Ball trajectory problems with initial velocities just became difficult of Calculus never existed.
Very nice
7:30 you don't need to be modest. 😰
Nice!
Good 👍
(3,0),(3,27),(infinity,27),(infinity,0) lol😂
Could you do a video on the differentiation of f(x) =|x|? I can't find an explanation for why it's x /|x|. (it makes sense but I'd like to know how you get there)
He does have it in some of his videos but I’m not sure which ones.
But the simplest way is rewrite |x| as sqrt(x^2) then differentiate the usual way, then replace the sqrt(x^2) back with |x| in the final answer. So
d/dx[|x|]
=d/dx[sqrt(x^2)]
=1/[2sqrt(x^2)] * 2x
=x/sqrt(x^2)
=x/|x|
The definition of absolute value is : |x|=sqrt(x^2)
If you want to differentiate |x| you can just take the derivative of sqrt(x^2) which is
x/sqrt(x^2)
= x/|x|
OK, thanks guys
|х| = х, if x≥0; -x if x < 0.
|x|' = x' if x>0, (-x)' if x
this explanation is more illuminating than the ones above
Without calculus he said
At 5:32 you divided the equation by 3 but you didn't multiply by 3 on either side to make the equality holds
Huh? No, he did not divide the equation by 3 at 5:32, what are you talking about? And there is no equality at 5:32, only an inequality. Are we talking about the same video???
He was just trying to figure out what to choose for a, b, c, and d. He was looking for a way to remove the x from the right side by choosing values for b, c, and d that would eliminate the -x from the (3 - x) he had for a. Because he chose those for b, c, and d, he then had to do that to both sides of his inequality. This way, he had a constant value on one side and then was able to proceed until he got his original equation on one side
using calculus is so much easier though.
It's always dumb to know the Calculus way and not use it. Especially in Physics or Mechanical Engineering or Drafting professional jobs.
@@lawrencejelsma8118well the point here is to explore and learn other possible ways to solve the problem instead of just looking for the easiest way. We never know if there may be a similar problem where using calculus may be much harder than this method, such as optimisation problems with 3 or more variables
How do you prove AM>GM. I suspect it might be .... calculus!
No, just algebra or geometry
en.m.wikipedia.org/wiki/AM-GM_inequality
I didn't understand that at all! And I have a Math A.A. ...I don't remember how to do it by calculus either. Embarassing!
Black magic
One of the conclusion in this video, calculus can save your time. Pls spend some time to learn it.
Newton saved humankind.
Doing it the calculus way shows that the maximum area (where the derivative = 0) occurs at x=3/4 (0.75) and not at x=9/4. I guess that the substitution should be reversed at the very end by dividing by 3.
"Doing it the calculus way shows that the maximum area (where the derivative = 0) occurs at x=3/4 (0.75) and not at x=9/4."
Actually, doing it the calculus way also gives x = 9/4. I don't know how you got x = 3/4. Could you please show your calculations?
"I guess that the substitution should be reversed at the very end by dividing by 3."
No, that makes no sense, why should one do that? Which "substitution" are you even talking about?
ahow is AM-GM proven? :DD
Can you prove AM-GM without calculus? I'd like to see how.
Yes, using algebra. I did that in this video (for the two variables case) th-cam.com/video/dCQeGIrkMWQ/w-d-xo.htmlsi=YnJTmzXxO0QMf4rO
I cheated and used calculus unfortunately!
👍
short method
we have to divide 3 in ratio 1:3
a - x = 3/4
x = 9/4
area = (3/4)(729/64)
= 2187/256
yea... prove it (or derive: where did the ratio 1 : 3 even come from)
@FaranAiki consider four positive quantities
(3 - x)/1 , x/3 , x/3 , x/3
with sum = 3
when we apply AM - GM inequality on these quantities we get the result.
in AM - GM inequality equality holds iff all quantities are equal hence
(3 - x)/1 = x/3 in other words ratio between (3 - x) and x is 1:3
and since their sum is 3 we can say that we have to divide 3 in the ratio 1:3
other examples -
(1) find the maximum value of (x^3){ (20 - x)^2 } if 0 < x < 20
we have to divide 20 in the ratio 3 :2
x = 12 , 20 - x = 8
maximum value is (12^3)(8^2) = 110592
(2) find the maximum value of (x^2)(y^3) if 3 x+2 y = 1 , x > 0 , y > 0
we have to divide 1 in the ratio 2 : 3
3 x = 2/5 , 2 y = 3/5 or
x = 2/(3)(5) , y = 3/(2)(5)
maximum value is 3/6250
(3) find the maximum value of xy if x+y = 8 , x > 0 , y > 0
we have to divide 8 in the ratio 1:1
x = 4 , y = 4
maximum value is 16
@raghvendrasingh1289 Ok so not a shorter way...
@@raghvendrasingh1289 Since all of that long explanation is necessary for solving the problem, your way was not shorter at all.
像我们国内中考题
ealry gang
太正了
First