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Mathematics
KenzeyL:

If 2x^2 + 3y^2 + 4z^2 = 12x 2 +3y 2 +4z 2 =1 and MM is the maximum value of 6x-3y-8z6x−3y−8z, evaluate \lfloor M \rfloor⌊M⌋.

offbeatsoldier33333:

i dont get it

KenzeyL:

Neither do I lol I just found it in my cousins math book

Convert:

This hurts my head- are you sure you have it written down exactly how it is?

KenzeyL:

Yes

Convert:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @KenzeyL Yes \(\color{#0cbb34}{\text{End of Quote}}\) So what are the random 2's doing there- and the `\lfloor M \rfloor⌊M⌋.`? You're telling me it's written like that in your cousins math book as you previously said?

KenzeyL:

I guess i don't really know i just wrote down what I saw

Convert:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @KenzeyL I guess i don't really know i just wrote down what I saw \(\color{#0cbb34}{\text{End of Quote}}\) Can you in any way, take a picture and upload it here?

KenzeyL:

I can't sorry. My light isn't on in my room and I can't have it on after 11

Convert:

Oh, that's all right but is the top part of this written as:\[2x ^{2}+3y ^{2}+4z ^{2}=12x ^{2}+3y ^{2}+4z ^{2}=1\]

Convert:

Because that's what I got from that-

KenzeyL:

uhm Yeah. I think so

Convert:

Well the \[3y ^{2}+4z ^{2}\] Would cancel out right? Because it's there on both sides.

KenzeyL:

Yes? I don't really know. Like I said this was my cousin's math book and she's out of highschool and I'm still in middle school. I'm just curious on this problem.

Convert:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @KenzeyL Yes? I don't really know. Like I said this was my cousin's math book and she's out of highschool and I'm still in middle school. I'm just curious on this problem. \(\color{#0cbb34}{\text{End of Quote}}\) Ohh alright so this isn't your work you're just wondering as to how it would be solved?

KenzeyL:

Yes

Convert:

Well I've been trying to figure it out and once you cancel out those terms it'd leave you with \[2x ^{2}=12x ^{2}=1\] Which wouldn't be possible because 2x^2 can't equal 12x^2 so I'd honestly have no clue as to how you would solve it.

KenzeyL:

Oh okay. Thanks anyway. Too many numbers for my head.

Convert:

And I think the second part was from another problem in which "l" floor would be divided by "r" floor in that equation.

Convert:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @KenzeyL Oh okay. Thanks anyway. Too many numbers for my head. \(\color{#0cbb34}{\text{End of Quote}}\) Yw 😂 and I know- I'm doing Geometry for my 8th grade class and it already confuses me enough lol

KenzeyL:

lol dang Ur only in 8th and you could almost solve that. Impressive. I suck at math.

Convert:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @KenzeyL lol dang Ur only in 8th and you could almost solve that. Impressive. I suck at math. \(\color{#0cbb34}{\text{End of Quote}}\) Eh well after you get past algebra 1 then it's the same thing for Geometry, that I know just with shapes 😂. And I'm sure your not bad at math, most people who say that just have bad teachers lol but a lot of people on here can help you understand something if your teacher doesn't explain it well enough to you in class and you don't get it.

KenzeyL:

Lol. My teacher is ok kinda. She makes us learn things on our own most of the time.

Convert:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @KenzeyL Lol. My teacher is ok kinda. She makes us learn things on our own most of the time. \(\color{#0cbb34}{\text{End of Quote}}\) Most likely why you think your bad at math lol, you just gotta see it in a different way.

KenzeyL:

Yeeah. I do NWEA in my school. And I got the highest language arts score in my grade so i personally think Language arts is more of my thing than math.

Convert:

Yeah we do NWEA for final exams and tests in classes, but it's not about what your good at that'll determine your job/hobby in the future, it's what you enjoy doing. Even if it's something you're not good at.

KenzeyL:

Well the job I want is a children's psychologist. And i don't really know if I need math or language arts for that lol

Convert:

Well for that you would most likely have to study neurology and the brain which would relate more to science and biology lol

KenzeyL:

Yeah most likely

KenzeyL:

Yo uh AZ you may wanna chill. It's ok if the problem ain't solved. I don't need a whole book. My head can't handle anymore bro. Thanks for the effort tho.

AZ:

Let M be the maximum value of \((6x−3y−8z)\), subject to \(2x^2+3y^2+4z^2 = 1\). Find [M]. So \( f(x, y, z) = 6x−3y−8z \) and it's subject to the constraint \(2x^2+3y^2+4z^2 = 1\) And we'd solve this using Lagrange multipliers \(\nabla f = \lambda \nabla g\) \((6, -3, -8) = (4\lambda x, 6\lambda y, 8\lambda z)\) \(4\lambda x = 6\) \(6\lambda y = -3\) \(8\lambda z = -8\) so we get \(x = \dfrac{3}{2 \lambda}\), \(y = -\dfrac{1}{2 \lambda}\), \(z = -\dfrac{1}{ \lambda}\) Let's plug that into our constraint \(2x^2+3y^2+4z^2 = 1\) \(2\left(\dfrac{3}{2 \lambda}\right)^2 +3\left(-\dfrac{1}{2 \lambda}\right)^2+4\left(-\dfrac{1}{ \lambda}\right)^2 = 1\) Solve for \(\lambda\) lots of steps involved, just used wolframalpha and got \(\lambda = \pm \dfrac{\sqrt{37}}{2}\) and let's use our positive \(\lambda\) values and we'll get \(x = \dfrac{3}{ \sqrt{37}}\), \(y = -\dfrac{1}{ \sqrt{37}}\), \(z = -\dfrac{2}{ \sqrt{37}}\) and if we plug it into the function \( f(x, y, z) = 6 \left(\dfrac{3}{\sqrt{37}}\right) - 3\left(-\dfrac{1}{\sqrt{37}}\right) -8 \left(-\dfrac{2}{\sqrt{37}}\right)\) \( f(x,y,z) = \sqrt{37} = 6.0827\) And similarly use the negative value for \(\lambda\) and plug it all in and follow through, you'll get \(x = -\dfrac{3}{ \sqrt{37}}\), \(y = \dfrac{1}{ \sqrt{37}}\), \(z =\dfrac{2}{ \sqrt{37}}\) And evaluating that results in \( f(x,y,z) = -\sqrt{37} = -6.0827\) So we know our maximum is at \(\left(\dfrac{3}{ \sqrt{37}}, -\dfrac{1}{ \sqrt{37}},-\dfrac{2}{ \sqrt{37}}\right) \)and if we use the floor function (which just gives us the highest integer value closest to x) we get our final answer ⌊M⌋ = 6

AZ:

Been ages since I've done this so would love if @jhonyy9 @Angle @Vocaloid could check over it (:

KenzeyL:

omg you did it.

Convert:

Wow AZ-..

AZ:

(knew that the asker really didn't need help with it so I just answered it in its entirety)

offbeatsoldier33333:

ive been looking at it for hours and still ont get it

jhonyy9:

\(\color{#0cbb34}{\text{Originally Posted by}}\) @AZ Been ages since I've done this so would love if @jhonyy9 @Angle @Vocaloid could check over it (: \(\color{#0cbb34}{\text{End of Quote}}\) perfect job and explained understandably - CONGRATS @AZ !

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