Question

How do I solve this? a, b, c, d, e, f are positive numbers, if a+b+c+d+e+f=6 and a^2+b^2+c^2+d^2+e^2+f^2=36/5 how do I find the largest possible value of a^3+b^3+c^3+d^3+e^3+f^3?

Answer 1

does this help

Answer 2

this makes no sense, and it’s wrong

Answer 3

im just giving him a hand-

Answer 4

you’re writing nonsense and pretending it answers the question when it isn’t related at all

Answer 5

The point E can be obtained by folding the square along the bisector of
angle BAC, so that AB is folded onto the diagonal with B landing on E. If F is
the point where the bisector intersects B C, then B F folds to E F. Thus E F -1 AC
and BF = EF. Since L.EFC = 45° = L.ECF, FE = EC.
l.1(b). WC! = IBC! - IBFI = IAEI - ICEI = 21AEI - lAC! and IEC! =
lAC! - IAEI = lAC! - IBC!.
l.3(a).
n Pn qn rn
1 1 1 1
2 3 2 3/2 = 1.5
3 7 5 7/5 = 1.4
4 17 12 17/12 = 1.416666
5 41 29 41/29 = 1.413793
1.3(c). Clearly, rn > 1 for each n, so that from (b), rn+1 - rn and rn - rn-I have
opposite signs and
1
Irn+l - rnl < 41rn - rn-II·
Since r2 > 1 = rl, it follows that rl < r3 < r2. Suppose as an induction
hypothesis that
rl < r3 < ... < r2m-1 < r2m < ... < r2.
Then 0 < r2m - r2m+1 < r2m - r2m-1 and 0 < r2m+2 - r2m+1 < r2m - r2m+J.
so that r2m-1 < r2m+1 < r2m+2 < r2m. Let k, 1 be any positive integers. Then, for
m > k, I, r2k+1 < r2m-1 < r2m < r21.
l.3(d). Leta be the least upper bound (i.e., the smallest number at least as great as
all) of the numbers in the set {rl, r3, r5, ... , r2k+l, ... }. Then r2m-1 ::::: a ::::: r2m- the point is its useless

Answer 6

im using real math tho-

Answer 7

Don’t you just love plagiarism

Answer 8

i cant-

Answer 9

don't you love it when someone copy pastes the solution to 1978 USAMO 1 instead of this problem

Answer 10

i really cant with yall- 🙃

Answer 11

I CANT-

Answer 12

please take the time to READ what you copy pasted. we're not dumb

Answer 13

Thick ya real smart saying I copy and pasted it?-

Answer 14

I cant-

Answer 15

well then actually solve the problem please

Answer 16

I Made A Joke.

Answer 17

I gave him wrong info for a reason

Answer 18

who is "him"

Answer 19

I said it already

Answer 20

The person?

Answer 21

Oh no-

Answer 22

bruh

Answer 23

IK YALL CANCELLED ME

Answer 24

-

Answer 25

did u use it..?

Answer 26

any actual help?

Answer 27

LOL.

Answer 28

Omg- I'm logging off for today.

Answer 29

Okay, so in order to make
\(\sf{~~~a^2+b^2+c^2+d^2+e^2+f^2}\)
true, we need to find out what mixture of values will get you “\(\sf{\dfrac{36}{5}}\)” that also equal 6.

So what values of a,b,c,d,e, and f, can not only get you “\(\sf{\dfrac{36}{5}}\),” but also get you 6?

Answer 30

i tried but i can't figure out how to make both equations true at the same time

Answer 31

Try remembering negatives and decimals, I know it widens the range, but yk

Answer 32

it says positive though

Answer 33

so I plugged in a=b=c=d=1 and solved for e and f. I got \(1\pm\frac{\sqrt{15}}5\), which makes a^3+b^3+c^3+d^3+e^3+f^3 equal to \(\frac{48}5\) but how would I prove that it is the largest possible value?

Answer 34

Go through the \(\sf{^3}\)s and solve it using the answers you got for the \(\sf{^2}\)s

Answer 35

how would i prove that it's the largest possible value though? it's possible there are different a, b, c, d, e, f that give a larger value

Answer 36

Unless- was that the answer you got for the 3rds-?

Answer 37

Well the possibilities of a, b, c, d, e, and f all have to equal 6 (as in the basic equation)

But using the \(\sf{^2}\)s and solving what the values of a, b, c, d, e, and f are is what will give you the value for the \(\sf{^3}\)s

Answer 38

If you get what I’m saying

Answer 39

there are infinitely many possibilities for a,b,c,d,e,f, how do i try all of them

Answer 40

That’s why the squares are there

Answer 41

there's still infinitely many though

Answer 42

For the first equation there is, but not if you solve for the variables

Answer 43

On the second equation*

Answer 44

if i plug in any values of a,b,c,d then there are two variables left and two equations which gives a unique solution for e and f, so how do I try all possibilities of a,b,c,d? they are real numbers so there are infinitely many possibilities

Answer 45

Look at it this way, what values can a b c d e and f equal that give you “36/5” when they are squared, but give you “6” when they are not squared?

Answer 46

infinitely many

Answer 47

You have to get an exact value of 7.2 though.

Answer 48

there are 6 variables but only 2 equations, there's not enough information to solve for the variables

Answer 49

You have to use the fact that a b c d e and f (their values themselves) all combine to equal 6 in every equation

36/5 is just the answer to what the variables all squared would equal

Answer 50

i found a few more: (0.8,0.8,0.8,0.8,0.8,2), (0.8,0.8,0.8,0.8,2,0.8), (0.8,0.8,0.8,2,0.8,0.8), (0.8,0.8,2,0.8,0.8,0.8),(0.8,2,0.8,0.8,0.8,0.8),(2,0.8,0.8,0.8,0.8,0.8), which all make the cubed expression 10.56, and (0,1.2,1.2,1.2,1.2,1.2), which makes the cubed expression 8.64

Answer 51

how do i prove that it is the maximum?

Answer 52

Okay so I looked back to your answer that you got from solving them, and there’s an easier way to prove this, if I’m remembering this correctly

The change between \(\sf{^1}\) and \(\sf{^2}\) is 1.2

Now, given that and the value you achieved, \(\sf{\dfrac{48}{5}}\), you can prove it

This is simply by understanding the equation has a straight increase, so you can use the amount of increase per powers to figure out the next \(=\)

It jumps from \(6\) to \(7.2\) (1.2 increase) and then to \(9.6\), which is a 2.4 increase

Meaning your equation is increasing at this straight incline

Answer 53

but I found a larger value with different a, b, c, d, e, f, which is 10.56

Answer 54

this is trivial

Answer 55

when did i ask

Answer 56

if it's trivial then what's your proof

Answer 57

ok I think I have a proof that 10.56 is optimal
first b+c+d+e+f=6-a and b^2+c^2+d^2+e^2+f^2=36/5-a^2 so by cauchy schwarz (36-5a^2)>=(6-a)^2, so 6a^2-12a<=0, which means a<=2. Now, we have \((a-4/5)^2 (a-2)\leq0\) so \(a^3-3.6a^2+3.84a-1.28\leq0\), so \(a^3+b^3+c^3+d^3+e^3+f^3\\\leq3.6(a^2+b^2+c^2+d^2+e^2+f^2)-3.84(a+b+c+d+e+f)+7.68\\=25.92-23.04+7.68=10.56\)
this means that the maximum value is \(\boxed{10.56}\) :)

Answer 58

nice -

Answer 59

u have time to do that but not google?

Answer 60

alecks bad

Answer 61

also oren sobad imagien only getting 35.7%