Question

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⌋.

Answer 1

i dont get it

Answer 2

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

Answer 3

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

Answer 4

Yes

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

Because that's what I got from that-

Answer 11

uhm Yeah. I think so

Answer 12

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

Answer 13

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.

Answer 14

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

Answer 15

Yes

Answer 16

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.

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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.

Answer 25

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.

Answer 26

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

Answer 27

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

Answer 28

Yeah most likely

Answer 29

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.

Answer 30

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

Answer 31

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

Answer 32

omg you did it.

Answer 33

Wow AZ-..

Answer 34

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

Answer 35

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

Answer 36

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