Question

\[\text{ Show } x^2y^2(4-x^2)(4-y^2) \le 16 \]
\[\text{ for all real numbers } x \text{ and } y \text{ such that } \\ |x| \le 2 \text{ and } |y| \le 2\]

Answer 1

I don't buy that for all x,y , is there a restriction?

Answer 2

changed the question to reflect the changes

Answer 3

kk lol

Answer 4

so if \[|x|\le2\] then \[|x|^2=|x^2|=x^2\le2^2\]

Answer 5

similar for y, yes?

Answer 6

ok

Answer 7

eh wait, maybe just use the difference of squares, what topic are you on in class atm?

Answer 8

Well can talk about any approach we want.

Answer 9

But I can tell you I got it from my special book of magical calculus spells.
(It is just james stewart calculus 6th edition)
I have one approach in mind but like to see what you guys can come up with as well. Like any way you guys can think of.

Answer 10

Even ways that would hurt my brain.

Answer 11

For the above, I was going to use a limit as it approaches x,y |4|
So second idea
\[x^2y^2(4−x^2)(4−y^2)≤16\]
could become: \[x^2y^2(2−x)(2+x)(2−y)(2+y)≤16\] so then we can say the maximum of each is obtained at 0 or 2, so we end up with a zero in the multiplication string when each individual attains it's max. which discredits those

Answer 12

but the smart thing would probably be to apply a derivative and determine that way

Answer 13

we would have to show that the derivative was never positive

Answer 14

we would have to use partial derivatives in a R3 environment, I think

Answer 15

\[ \text{ Let } f(z)=z^2(4-z^2) \text{ where we have } -2 \le z \le 2 \]
\[f(\pm 2)=0\]
\[f'(z)=4z(2-z^2)\]
\[f'=0 \text{ gives } z=0 \text{ or } z=\pm \sqrt{2}\]
\[\text{ but } f(0) \text{ doesn't yield our max value }\]
\[f(\pm \sqrt{2})=4 \text{ which is the max value }\]
\[ \text{ So } f(z) \le 4 \]

\[f(x) \cdot f(y) \le 4 \cdot 4 =16\]

Answer 16

Is the way I was thinking

Answer 17

but i want to see the partials too

Answer 18

I don't recommend using z, make it something else there. but is that the derivative? let me see...

Answer 19

also, based on your given, you would have that expression squared

Answer 20

but we don't know that x=y

Answer 21

i feel like we would need to set it up like this in 3 space(maybe, \(\color\orange{\text{IE if you know how to do this just substitute yourself into the game }}\))

Answer 22

no... i can't think of how to parameterize it... hmmm

Answer 23

"I don't recommend using z, make it something else there. but is that the derivative? let me see...
"
why is this?
"also, based on your given, you would have that expression squared
"
what expression?

-
also x and y do not have to be equal but f(x) and f(y) will still both have max value 4

Answer 24

on the intervals from x=-2 to x=2
and from y=-2 to y=2 that is

Answer 25

so partials, would yield:
\[x^2y^2(4−x^2)(4−y^2)≤16\\
f_x=y^2(4-y^2)(2x(4-x^2)+x^2*-2x)\\
f_y=x^2(4-x^2)(2y(4-y^2)+y^2*-2y)\\f_z=0\]

Answer 26

and z implies complex plane alot(a heck of a lot for me lately) so I try not to use it for real numbers anymore so I don't comfuse myself on roots

Answer 27

the expression was your change of variable to Z, I assumed you let x and y be the same in which case your z expression would have been squared

Answer 28

what about switching to polar?

Answer 29

I didn't really need to introduce another variable
I could have found the max values for f(x)=x^2(4-x^2) on x<= 2 and f(y)=y^2(4-y^2) on y<= 2
but I thought doing the same thing twice was a waste of time

Answer 30

|x|<=2 *
|y|<=2*

Answer 31

https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=graph%20z%3Dy%5E2(4-y%5E2)x%5E2(4-x%5E2)

Answer 32

yea... hmmm

Answer 33

hmm
x^2y^2<=16
(4-y^2)(4-x^2)<=16
but the relation btw those two terms is inverse xD
like maximize x^2y^2 ,means minimize (4-y^2)(4-x^2) ...ect
we need some cool theorem hmm

Answer 34

yea, I mean they cancel each other out essentially

Answer 35

they dont cancel ...

Answer 36

somehow , we need to consider cases less than 16 :)

Answer 37

*

Answer 38

you know you guys consider the max values on the intervals from -2 to 2
for x^2 and (4-x^2) separately
but why can't we consider the max value of the product of x^2 and (4-x^2)

Answer 39

how do we find that max value, optimization

Answer 40

well that is what I did in my post above

Answer 41

i feel like the max should be 3

Answer 42

ok here is something since bounds are equivalent
i'll consider
|x|=|y|

x^4(4-x^2)^2
and for less values assume
x^4=s

x^4(16-8x^2+x^4)
16x^4 -8x^6+x^8

s^2-8s^(3/2)+16s

Answer 43

well, 9

Answer 44

eh i'll see now if that works

Answer 45

\[f(x)=x^2(4-x^2) \text{ on } x \in [-2,2] \\ f'(x)=2x(4-x^2)+x^2(-2x)=-4x^3+8x=-4x(x^2-2) \\ \\ f'=0 \text{ gives } x=0 \text{ or } x=\pm \sqrt{2} \\ \text{ so plug in our criticqal numbers along with our endpoints into } f\]

Answer 46

to see what would be our max value

Answer 47

f(-2)=0
f(2)=0
f(0)=0
f(-sqrt(2))=4
f(sqrt(2)=4

Answer 48

our max value is 4

Answer 49

for f(x)=x^2(4-x^2)

Answer 50

so our max value for f(y)=y^2(4-y^2) is 4

Answer 51

ok it works
http://www.wolframalpha.com/input/?i=maximize%28s%5E2-8s%5E%283%2F2%29%2B16s%29

Answer 52

ok so at sqrt of 2 yea

Answer 53

I'll be back later to check, I have an assignment due at midnight

Answer 54

thats it ? xD

Answer 55

well i mean i'm interested in other ways if there are other ways

Answer 56

there is L M theorem but im too tired to remember it xD

Answer 57

L M?

Answer 58

but then i thought its only polynomial

Answer 59

lagrange multiplier ?

Answer 60

yes let me see ifi found my books

Answer 61

wait not yes lagrange sorry

Answer 62

yes when u said L M , so i replied yes

Answer 63

that product of functions thingy is a neat trick @myininaya :)
finding critical points isn't that complicated here as everything is factored already

Answer 64

ok cant find it anywhere XD

Answer 65

\[f(x,y)=x^2y^2(4-x^2)(4-y^2)\]

\(f_x = y^2(4-y^2)(8x-4x^3)\)

we dont need to work the other partial as we can use symmetry

\(f_x = 0 \implies y = 0 , \pm 2\), \(x = 0, ~x = \pm \sqrt{2}\)

the function feels continuous every where in the given domain so it has to achieve maximum value at critical points or the boundary points

Answer 66

hmm

Answer 67

why do you want to use lagrange multipliers here ?

Answer 68

oh we can interpret the given bounds as constraints is it ?

Answer 69

no one wanna use it

Answer 70

do you guys just want to do another problem?
or do you guys have fun problems you can post?

Answer 71

if u have post

Answer 72

ok I will just choose a random one

Answer 73

Alright, lagrange multipiliers makes no sense to me here.. .i admit im confused a bit about seeing the difference between "bounds" and "constraint function"

Answer 74

il stick to the modest way of finding critical points or the other two methods that were discussed in this thred :)

Answer 75

wat modest ?

Answer 76

what other two way

Answer 77

1) myininaya's method of taking product of max values of two functions of different variables :
http://gyazo.com/eca9604182d11a282ad7f7e235a4f898

2) your method which you know :)

3) finding critical values

Answer 78

i consider last method is modest because thats the method textbooks teach and also thats the method you turn to when your brain stops working :P