Finding the BIGGEST rectangle under y=x^3 (but NO calculus!)

We will solve a typical calculus optimization problem without using calculus derivative. The goal is to find the biggest rectangle under y=x^3 from x=0 to x=3 and we can actually use AM-GM inequality to solve this!
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[remigeron]

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).

[pwmiles]

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

[cdkw]

Nice method but I'd still use calculus though!

[peterbrockway]

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.

[ronbannon]

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.

[andrec.]

Na minha opinião, um dos melhores da Internet: objetivo, claro e resolve problemas valorosos! Obrigado, companheiro!

[nocturnalvisionmusic]

Calculus master is back with more awesome stuff 🙌🤗

[graf_paper]

Oh man that is a satisfying method. Bravo.

[abcdwxyz]

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

[gregorymagery]

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

[kenhaley]

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.

[skilz]

Beautifully done, what a piece of artistic work.

[HeManGNichtDualismus]

why would am gm come in your head in the first

[aranarus]

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.

[mathemagician]

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

[lucsas]

i started trembling and sweating at the no calc statement

[mathemagician]

“Without calculus” proceeds to use a result from Real Analysis without justification 😂

[jamescollier]

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

[colbyforfun]

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?

[paytonholmes]

How is the geometric mean the maximum? Please explain.