Calculus optimization: given surface area, want largest volume of an open-top box
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- Опубликовано: 9 сен 2024
- Calculus optimization! Given the surface area, want the largest volume,
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You are quite literally my hero
x² + 4xy = 1200
4xy = 1200 - x²
y = (1200 - x²)/4x
Maximise x²y:
d/dx x²y
= d/dx x²(1200 - x²)/4x
= d/dx 300x - ¼x³
= 300 - ¾x²
to maximise, set to 0 to find critical value:
= 3(10 - ½x)(10 + ½x) = 0
x = ±20
x = -20 doesn't make sense for a physical object, so let us take the positive value:
x = 20 units.
y = (1200 - x²)/4x
= (1200 - 400)/80
= 800/80
y = 10 units.
volume = x²y = 20·20·10 = 4000 cubic units.
This is 4th time of AM-GM. Let x^2 + 4xy = a^2 be fixed. We note
x^2 + 2xy + 2xy = a^2.
By AM-GM inequality we have
( x^2・2xy・2xy )^( 1/3 )
there are more analytical problem in application of derivatives than the one which ur solving 😊.....ur making it look so easy ...
Tomorrow is mature exam in basic maths and 11.05 is extended maths in Poland. This is typical optimisation question in extended exam for 7 of 50 points
Can u make another video where u solve this kind of problem using anadelta? It would be great!
P.S:Nice video once more!!!
How would you do it if the y value wasn’t constant? Like one side was taller than the other? The problem didn’t specify a constant height. Could you maximize volume more?
In that case I think you'd need to develop general equations for the volume and surface area, since we wouldn't be able to take the shape to be a rectangular prism. Y itself would have to be some function in terms of X, so I feel like it would start to become more like a 3D analytical geometry problem. Would love to see it attempted though, it sounds like it could be an interesting process!
Hi! I recently watched one of your older videos about the summation of 1/n^3 as n goes from 1 to infinity. By trying on the calculator, i think, I found a nice form of that approximation for the sum. It's 1,7/sqrt(2). What do you think? :)
Setting
a := Σ_{ n = 1 }^{ ∞ } 1/n^3
the calculator shows that
a = 1.20205690…
1.7/√2 = 1.20208152…
How much error is between a and 1.7/√2 by your method ?
@@user-dl8nk5bf8v Around 0,00002463.
@@danielkovacs6809
I don't mean that. How did you get 1.7/√2 ? No method but only using the calculator ?
Ok so if the box has an open top and a square base let’s set the variables. Let’s say that a is the side of the square. So we’ll have 4 other rectangles let’s say their heights are b. So a^2 + 4ab = 1200. And the volume will be a^2 * b. Now I believe we put it into 1 variable and plug it into the volume function. So ig I’ll solve for b and later I’ll derive with respect to a. So factor: a(a+4b)=1200 and divide by a: a+4b=1200/a. Now I subtract and divide I think. 4b=(1200/a)+a and now: b=((1200/a)+a)/4. Ok so the function (once simplified) will be (let’s say V(a) is the volume) V(a)=(a^3)/4 + 300a. Ok and the derivative is just V’(a)=(3a^2)/4 + 300. Now we set it to 0. 0=(3a^2)/4 + 300. We can now solve for a. -300=(3a^2)/4 and now -400=a^2. Uh oh. Well I messed up. I should’ve chose b maybe or i did smt wrong along the way.
Lemme redo this
Ah my mistake was I added a instead of subtracting
Edit: ok here we go
4ab=1200-a^2
b=(1200-a^2)/4a
V(a)=a^2 * ((1200-a^2)/4a)
And ig I’ll foil
V(a)= (1200a^2 - a^4)/4a
And simplify again…
V(a) = 300a - (a^3)/4
And derive
V’(a)=300 - (3a^2)/4
And set to 0
0=300 - 3/4 * a^2
And subtract
-300= -3/4 * a^2
And divide by 3/4
-400 = -a^2
And now I can multiply by -1 and sqrt
20 = a
And finally I can find b and the volume
b=(1200-a^2)/4a
So b= (1200 - 400)/80
And b= 800/80
Here we go: b = 10
10*20*20=4000
So the largest possible volume for the box is 4000cm^3
Edit again: I’ve made a mistake but I learned from it and I found it
Edit: forgot about the negative square root but I don’t think it matters in this bc I’m pretty sure in the real world that boxes don’t have negative volumes and dimensions lol
How is this comment is 1 day ago
sorry im about to unsubscribe if there is uploaded another cube the box question video. do something like polynominal regression with an alpha smoothing