Question

If a,b,c, d and p are positive such that a+b+c+d + p= 8, a^2 + b^2 + c^2 + d^2 +p^2 = 16, then the maximum value of p is

Answer 1

a,b,c,d, and p are not necessarily natural numbers ; they are positive real numbers.

Answer 2

@ganeshie8 @hartnn

Answer 3

can they be zero ? and real not integer /?

Answer 4

They cannot be zero.

They are real but not natural numbers.

Answer 5

ummm

Answer 6

:)

Answer 7

they are positive and not 0
so 4^2 = 16 which makes it 1 number
so others are less than 4

Answer 8

So this is a constrained optimization problem.
non-linear
thinking...

Answer 9

so 3^2 = 9,2^2 =4 and rest would be 1 1 1

Answer 10

so maximum value would be 3
dont u think

Answer 11

aa they can be R not N

Answer 12

oh

Answer 13

but at the very least 3 is a lower bound for p

Answer 14

I am going to make the following argument. I am not actually certain that it's strictly speaking rigorous, but I think it will solve the problem here. Let's find out anyway.
The argument is that there's no difference in the equations between a, b, c, and d. They all appear with the same constraints, same signs, and same functionality in the equations. So I argue that you can aggregate them together

Answer 15

Maximize : \(f(b,c,d,p) = p\)
subject to :\([8-(b+c+d + p)]^2+ b^2 + c^2 + d^2 +p^2 -16 = 0\)

Answer 16

well

Answer 17

lagrange multipliers gives the fast answer

Answer 18

\[\langle 2[8-(b+c+d+p)] +2b, 2[8-(b+c+d+p)] +2c, \\ 2[8-(b+c+d+p)] +2d, 2[8-(b+c+d+p)] +2p\rangle \\= \lambda \langle 0,0,0,1\rangle \]

Answer 19

I see lots of symmetry in the problem - we may use it if we can figure out a way - but above method gives you answer directly ^

Answer 20

I'd probably say use cauchy–schwarz inequality for a,b,c and d

Answer 21

have you tried it @vishweshshrimali5 ?

Answer 22

We start with the two constraints given.
We have positivity a,b,c,d,p must be in the positive reals (excluding zero).
Furthermore
a+b+c+d+p = 8
a^2+b^2+c^2+d^2+p^2 = 16
What is the maximum that we can make p? Given this system.
From the first equal and the positivity constraint, we can see that

p = 8 - a - b - c - d. Since a,b,c,d>0. It follows that p < 8. So this is just a sanity check on any answers we may get.

Now the symmetry argument because it's obviously hard to handle the four variables here. (We cannot use simplex here because we have a non-linear constraint a^2+...+p^2 = 16)

The symmetry argument says, well, a,b,c,d have the same sign and make the same contribution to the constraint. So let's reduce them to a single variable "x" multiplied by 4.
If we do that, we get a 2 variable system that we can solve.

p = 8 - 4x
p^2 = 16 - 4x^2
substitute p into the second equation
(8-4x)^2 = 16 - 4 x^2
Expand
64 -64x + 16x^2 = 16 - 4x^2
Bring everything to one side
20x^2 - 64x + 48 = 0.
Divide out GCF of 5, apply quadratic formula

x = 16 +- Sqrt[16^2-4*5*12] all over 10
This gives x = 2, x = 6/5

x = 2 is not what we want because it makes p = 0
which violates the positivity constraint
So we take x = 6/5
This yields p = 16/4 from p = 8 - 4x

We check consistency of our answer to make sure that we did not mess up and violate the non-linear constraint 4x^2+p^2 = 16. We didn't. Okay, we have found the answer.

Answer 23

Well, if you argue that "a" "b" "c" and "d" are essentially 'the same' in the equation. Namely "x" then a^2 = x^2, b^2=x^2, and so a^2+b^2+c^2+d^2 = 4x^2

Answer 24

@Miracrown , your answer was absolutely correct (the answer should be 16/5).

Here is how I got the answer,

\[\large{(\cfrac{a+b+c+d}{4})^2 \le \cfrac{a^2 + b^2 + c^2 + d^2}{4}}\]

Answer 25

ah... perhaps...

Answer 26

So the thing is. If you make "a" and little greater and "b" a little smaller in the hopes that you improve you value of "p" What would that gain you? Nothing it's the same as smoothing out the change over a single variable "x"

Answer 27

A Mathematica constrained optimization solution is attached.

Answer 28

according to cauchy–schwarz inequality\[4(a^2+b^2+c^2+d^2) \ge (a+b+c+d)^2\]using 2 constraints\[4(16-p^2) \ge (8-p)^2\]it gives\[p(5p-16) \le 0\]and finally\[0 \le p \le \frac{16}{5}\]

Answer 29

wow, that is crazy cool how many ways there were to solve this thing! :)

Answer 30

Because a, b, c, d don't have a different coefficient in the constraints
If they had different coefficients, it would be an entirely different and much harder problem

Answer 31

Yeah @mukushla , that's what I used.

Answer 32

Great work @mukushla and @Miracrown

Answer 33

probably Cauchy–Schwarz inequality is the best bet for the problem

Answer 34

3.2 close enough *wears glasses*

Answer 35

lol @aajugdar

Answer 36

could someone briefly explain how Cauchy-Schwarz applies? I've seen it used with dot product for the most part, is that hidden in here somewhere?

Answer 37

The thing is. There's no general method to solve these.
It's just cleverness.

Answer 38

If you have systems of linear constraints, then you can use simplex
But as soon as you get non-linear... it becomes tricky

Answer 39

So, cauchy-schwarz is basically the triangle inequality
Do you know it by this name? @jtvatsim

Answer 40

sure, |A + B| <= |A| + |B| right?

Answer 41

Yes, for the "2D" case

Answer 42

ah, so it generalizes then.

Answer 43

exactly

Answer 44

got it, so |A+B+C+...|^2 <= (|A|+|B|+|C|+...)^2

Answer 45

yes

Answer 46

So if a higher dimensional space, it's a longer length to go to A,B,C individually, then in some sense the straight line path from A--> X

Answer 47

lagrange is the general method right @Miracrown

Answer 48

ok, that seems reasonable that our notion of distance carries over

Answer 49

@rational Ah, you mean, if you want to use Langrangian multipliers? Actually, I did not think about that, but, yes, you could use them here

Answer 50

@jtvatsim yes

Answer 51

So cauchy-schwarz also guarantees tightness because, so you know that the distance can get no right than 4(a^2+b^2+c^2+d^2)

Answer 52

That's important, otherwise, you could have some really high upper bound which does not given you a meaningful constraint on p

Answer 53

ok, I think I understand everything except where the 4 came from, is that just coming from the fact that we have 4 variables?

Answer 54

thanks for taking the time to explain this much of the details :)

Answer 55

No problemo! Well... basically, but I'll explain

Answer 56

(sum i=1 to n of a_i*b_i)^2
must be <= (sum i =1 to n of ai^2) * (sum i = 1 to n of bi^2)
This is the cauchy schwarz inequality phrased in terms of sequences of real numbers
so
if we let b = sequence of 1s
(four ones)
and a = sequence of our four variables
then you see the 4 naturally comes from adding 1+1+1+1 for bi^2
thus (a+b+c+d)^2 <= 4*(a^2+b^2+c^2+d^2)
That's where the 4 comes from

Answer 57

ok, I'm copying and pasting this for further review. :)

Answer 58

(Actually this is the context of the first discovery of the cauchy-schwarz inequality in terms of sequences of real numbers) This is it's original form, I believe. Now-a-days, we mostly use it when dealing with vector spaces and such

Answer 59

right, I usually think of vectors right away

Answer 60

ah, great I see based on the sums that that would naturally follow. thank you!

Answer 61

you're welcome! :)