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⌋.

i dont get it

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

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

Yes

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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?

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

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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?

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

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

Because that's what I got from that-

uhm Yeah. I think so

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

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.

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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?

Yes

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.

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

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

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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

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

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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.

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

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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.

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.

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.

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

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

Yeah most likely

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.

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

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

omg you did it.

Wow AZ-..

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

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

\(\color{#0cbb34}{\text{Originally Posted by}}\) @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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