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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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