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!
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
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 ?
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
sorry im about to unsubscribe if there is uploaded another cube the box question video. do something like polynominal regression with an alpha smoothing
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.
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!
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!!!
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 ?
@@田村博志-z8y 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