Question

HELP REALLY NEEDED!!!!!!!!
SEQUENCES AND SERIES
Arithmetic and Geometric Progressions
A.Mean and G.Mean Inequality!

Answer 1

@aaronq @agent0smith @allopersonwhat @AllTheLonelyKillers @BulletWithButterflyWings @BAdhi @bhaskar.pathk @BeckieBird @Bladerunner1122 @campbell_st @dragons1415 @e.mccormick @Frostbite @Figureskater120 @jigglypuff314 @thomaster @Tyler1204 @kaylarocks001 @iceicebaby @pymn @RadEn @NoelGreco @Loser66 @lulu22 @AkashdeepDeb

Answer 2

@hartnn

Answer 3

Pardon me for so many tags!!

Answer 4

really need help!!

Answer 5

sorry, i dont no that stuff :c

Answer 6

np:)...

Answer 7

for all of u..........this is wat wolfram gave me.......
still dint get it!!
If u can..Pls explain dat!!
http://math.stackexchange.com/questions/351976/finding-the-minimum-of-n-fraca3ca2bc-frac7a6b3cab2c

Answer 8

a minimum value is when all the partials are equal to zero ... baring any sort of saddlebacking of course

Answer 9

or at least that provides critical points of interest

Answer 10

and by partials u mean???

Answer 11

derivatives ....

Answer 12

kay..

Answer 13

http://math.stackexchange.com/questions/351976/finding-the-minimum-of-n-fraca3ca2bc-frac7a6b3cab2c
similar type.pls explain

Answer 14

Explain how to rewrite the equation using A,B,C.........

Answer 15

dunno about the method on that site ... hard to read

Answer 16

hmmm....

Answer 17

You can just start dividing each by the variables and imposing conditions. Eventually, you'll get it.

Answer 18

a bit explanatory
Pls @primeralph
type the equation and show pls!

Answer 19

just to practice my patials ....
\[F(a,b,c)=\frac{a+3c}{a+2b+c}+\frac{4b}{a+b+2c}-\frac{8c}{a+b+3c}\]
\[F_a=\frac{2(b-c)}{(a+2b+c)^2}-\frac{4b}{(a+b+2c)^2}+\frac{8c}{(a+b+3c)^2}=0\]
\[F_b=-\frac{2(a+3c)}{(a+2b+c)^2}+\frac{4(a+2c)}{(a+b+2c)^2}+\frac{8c}{(a+b+3c)^2}=0\]
\[F_c=\frac{2(a+3b)}{(a+2b+c)^2}-\frac{8b}{(a+b+2c)^2}-\frac{8(a+b)}{(a+b+3c)^2}=0\]
see, now its simple lol

Answer 20

lol

Answer 21

i spose getting like denominators and such might prove a task as well?

Answer 22

might

Answer 23

with the help of the wolf ...
\[\frac{a^3+6 a^2 b+13 a b^2+3 a b c-3 a c^2+8 b^3+15 b^2 c-13 b c^2+2 c^3}{a^3+4 a^2 b+6 a^2 c+5 a b^2+17 a b c+11 a c^2+2 b^3+11 b^2 c+17 b c^2+6 c^3}\]

Answer 24

OMG........O_O<faints for a micro sec>

Answer 25

\[F_a:3a^2+12ab+13b^2+3bc-3c^2=0\]
\[F_b:6a^2+26ab+3ac+24b^2+30bc-13c^2=0\]
\[F_c:3ab-6ac+15b^2-26bc+6c^2=0\]

oy vey thats still messy

are you expected to do this all by hand?

Answer 26

I guess.Its in the objective section!!!lol

Answer 27

obscene section maybe ....

Answer 28

lol

Answer 29

ans is C

Answer 30

Not sure man... :/
How have you done previous questions like this?

Answer 31

key gives that much..

Answer 32

Me........I am newb 2 this

Answer 33

@wio
@campbell_st

Answer 34

that's the first question..Imagine my condition for the entire job!!

Answer 35

guess I'll just have to live with this horror ........FOREVER!!

Answer 36

Or just use partial derivatives.

Answer 37

how???

Answer 38

Write each one as a function of the other and find the minimum values of the numerators and maximum of the denominators, or there about. Impose restrictions that make the values tend decreasing. .