any body can?

[phanbachloc]

:o can you find out the max and min of y
y=/a+2cosx/+/a+2sinx/ (/ / is abs)

[Gordon Bennett]

Well you can't really do that, since there are an infinite number of values, for both the max and the min, since it's a function of sinx and cosx. Specifying a range might help...

[pgrimes]

I think this is right...


The max value that cosx or sinx can have is 1, since functions fluctuate between 1 and -1. Since you are taking absolute values... cosx and sinx have a minimum value of 0.

So...

Ymax = (a + 2(1)) + (a + 2(1))
= (a + 2) + (a + 2)
= 2a + 4

Ymin = (a + 2(0)) + (a + 2(0))
= (a + 0) + (a + 0)
= 2a + 0
= 2a


your max/min y value is dependent on 'a'.

So, there you have it, you can't determine a max/min value in terms of a constant. But, in variable terms:

Ymax = 2a + 4
Ymin = 2a

[bugzpodder]

however, if a is -100 for example, it changes the answer. the maximum would no longer be 2a+4. actually it was never 2a+4. when cosx is 1, sin x can't possibly be 1 at the same time.

if you just ask whats the maximum value of y the equation can take on, the answer is that there is no maximum. the min value would be 0, when cosx=sinx and a=-2cosx (so one of the answers would be x=45 degrees)

[Destined Soul]

can you find out the max and min of y
y=/a+2cosx/+/a+2sinx/ (/ / is abs)Well, some simplification brings this to:

y = 2|a| + 2|cosx + sinx| (edit: a bit hard to justify, but it's somewhat correct)

It can be shown that cosx + sinx reaches a maximum where sinx = cosx (x = Pi/4 or 45 degrees) Cos(Pi/4) = 1/Sqrt(2)

So

y = 2|a| + 4/Sqrt(2) = 2|a| + Sqrt(8)

I think my math is right. :D

Destined

[bugzpodder]

you can't just take the "a" out of the absolute value.

abs(5-2) and abs(5)+abs(-2) is not the same thing!!

[bugzpodder]

even with ur edited post, it is still won't work. you seemed so sure. i'd be interested to see your justification.

[Destined Soul]

I suppose your somewhat correct. What needs to be done is for this to be broken into two cases:

a >= 0:

y = 2a + 2( |cosx| + |sinx| )

a < 0:
let b = -a
y = 2b + 2( |cosx| + |sinx| )

These are the same as

y = 2|a| + 2(|cosx| + |sinx|)

I, actually, was wrong in seperating both, but you CAN solve for the max value, which is y = 2|a| + |sqrt(8)|

Destined

[bugzpodder]

i told you, when a=5, and sinx=-1, it won't work!

[Destined Soul]

How won't that work? You'll get y = 12.

However, if a = 5, the max value will be about 12.83 (where x = (2n+1)*Pi/4, n is any integer)

If you think my sol'n isn't correct, please be a little more specific where I went wrong.

Destined

[bugzpodder]

lol which equation did u plug into??

if u plug into the given equation, you'll get 8, not 12!

[Destined Soul]

Originally posted by bugzpodder
i told you, when a=5, and sinx=-1, it won't work!

y = 2|a| + 2(|cosx| + |sinx|)

So plugging these in:

Note if sinx = -1, cosx = 0

y = 2|5| + 2(|-1|) = 2*5 + 2 = 12

Destined

[bugzpodder]

thats the equation you made up. the original equation is:
y=abs(a+2cosx)+abs(a+2sinx)

abs(5-2)+abs(5)=8!!

[Destined Soul]

Ok, so I assume something. Sadly, that's my downfall. I tend to see what I need and ignore the math that I probably shouldn't.

However, this is still solvable. Just not while I'm at work. :p Have to actually do stuff, you know? ;)

I'm pretty sure that the solution occurs when sinx = cosx. I still might be right, but I'm going to want to look this over better. Give me about an hour (I should be home around then). ;)

Destined

[Martin Wilson]

Max at 2|a|+2sqr(2) i.e. 45 degrees
Min at 2|a| + 2 i.e. 0 degrees

[phanbachloc]

:D i'm sorry,i make a mistake with a,a must be 1,now can you help me?and i got this problem:
sqrt(x-2)+sqrt(4-x)=x^2-6x+11
can u find x?it's very very difficult

[Gordon Bennett]

Standard procedure for finding max/min:

y = |a + 2cosx| + |a + 2sinx|

y' = 2.|a + 2sinx|.cosx - 2.|a + 2cosx|.sinx = 0

x = pi/4 or x = pi/4 - 1
y = 2|a| + 21.5 or y = 2|a| - 21.5

So if as you say a is always 1, the max and min of y are:
MAX: 2 + 21.5
MIN: 2 - 21.5

And that's correct as far as I can see.

[Gordon Bennett]

One more thing:

|a + 2cosx| + |a + 2sinx|

can't be simplified to:

2a + 2|cosx + sinx|

It should be:

2|a| + 2|cosx + sinx|

If you think about it, let's say a = -25
2a = -50 but 2|a| = 50: they're different.

[bugzpodder]

lol, when a=5, and cosx=-1, your formula won't work either Gordon. it won't work when a=1 either

[bugzpodder]

to phanbachloc


there are no solutions to the question u posed. its took me 2 min so its pretty easy

the solutions are suppose to be found in this equation:
x^2-6x+7=0
(x+1)(x-7)=0

x=-1, or 7

but the original question requires 2<=x<=4, so no solutions

[Destined Soul]

Originally posted by bugzpodder

the solutions are suppose to be found in this equation:
x^2-6x+7=0
(x+1)(x-7)=0Um, how did you get the LHS = 0? This would require Sqrt(x-2) + Sqrt(4-x) = 0.

So for x = -1, Sqrt(-3) + Sqrt(5) = 0?
And for x = 7, Sqrt(8) + Sqrt(1) = 0?

Both of which are impossible, even including the imaginary numbers, so I'd say these are not solutions.

Destined

[bugzpodder]

noo the last term of RHS is +11! i don't just make the LHS vanish you know, you should know that i am a little more advanced that do half a question! lol!

but seems that i've made a technical mistake, although its not as big as you originally thought


let me walk you thru: lets agree on the domain 4>=x>=2

sqrt(x-2)+sqrt(4-x)=x^2-6x+11

SBS

x-2+4-x+2sqrt((x-2)(4-x))=x^2-6x+11
2+2sqrt(-x^2+6x-8)=(x^2-6x+11)^2

originally i made a mistake that the RHS was not squared (i was doing it in my head) but now a double check says it is

so let -x^2+6x-8=y^2

so we have 2+2y=(-y^2+3)^2
2+2y=y^4-6y^2+9

y^4-6y^2-2y+7=0

now, looking at -x^2+6x-8=y^2, the maximum LHS can be is 1 (the vertex of the prabola lies on the line of symmetry, which is the average of the roots) the minimum is of course 0, since we want y^2>=0

so i quick check (factor theroem) says y=1 is a factor of the quartic.

so it factors into (y-1)(y^3-y^2-5y-7)=0
newton's estimating root way says there is a root between 3 and 4. that makes 2 real and 2 imaginary roots (or it may be 4 real roots).

I don't think there is another root between 0 and 1. a graph of that will obviously help. or those hard core guys can use the cubic formula for the equation to find the other 3 roots exactly. i am going to settle with y=1, which says x=3.

[phanbachloc]

:D in your posts,you say that there is no solution to my equation,
but actually,there is,if you give up,tell me ,i'll give you the solution that is very interesting and intelligent way!!i tell the truth

[bugzpodder]

lol yea i made a mistake saying no solutions. i found a solution (Refer to the post thats two above this one) saying that

x=3 satisfy your equation. I'd be interested in looking at your intelligent solution and compare it to mine though

[phanbachloc]

:) my way can prove that your cubic equation is not valid.it is:
sqrt(x-2)+sqrt(4-x)=x^2-6x+11
*let y=sqrt(x-2)+sqrt(4-x),2<=x<=4
=> y'=1/2sqrt(x-2)-1/2sqrt(4-x)
we see y'=0 when
sqrt(x-2)=sqrt(4-x)
=> x=3
so y(3)=2 ; y(2)=sqrt(2) ; y(4)=sqrt(2) (*)
from(*) we see max of y is 2.
*let f=x^2-6x+11=x^2-2*3x+9+2=(x-3)^2+2 (**)
from(**) we see min of f is 2
We have:
max of y =2 and min of f =2
=>y=f when y=2 and f =2
after solving we have x=3
do u agree with me?
by the way ,can you tell me your age,name ,i wanna be your friend?

[bugzpodder]

i agree with your analysis, but hey we got the same answer don't we? you mean my solution is not as good as yours lol. well to tell you the truth, I usually stick with elementary mathematics but in this case Calculus is certainly better and the solution you provided is pretty sweet (i thought at first it was just another boring algebra problem)

I think i have a way to avoid calculus:
in your argument, at the * you said that we see the max of y is 2 from .... I think u can convert LHS to what I did say its a maximum is to say that equation is a quadratic, and the max (since coefficient of x^2>0) of the quadratic lies on the line of symmetry (which is average of the roots). and same reason should be provided also for the minimum, either indicate that it is a quadratic and the co-efficient in front of x^2 is positive. or u can also simiply do the 2nd derivitive test for both cases although it can be a pain to take the second derivitive of the first function.

also, you used y=sqrt(x-2)+sqrt(4-x),2<=x<=4
so you should stick to that, and shouldn't use y(2)
if you want to use the function notation you can try something like f(x)=sqrt(x-2)+sqrt(4-x)


hey u can call me Bugz Podder, I am 15, and if you want, you can email me at bugz_podder@yahoo.com

[Masterbandit666]

sorry!!! sqrt(x-2)+sqrt(4-x)=x^2-6x+11
can u find x?it's very very difficult

Fairly obviously the solutions to this are x = 3 or x = 3 (repeated roots indicating that "y = (x - 2)0.5 + (4 - x)0.5" is a tangent to the line "y = x2 - 6x + 11" at the point (3, 2)).

Just x = 3 (what most people seem to have got) is adequate for a purely algebraic equation, but you didn't specify the context, and if the context was the intersection of two curves then the fact that it's a repeated root (i.e. x = 3 or x = 3) is quite important, as the repeated root proves they are tangential to one another.

[phanbachloc]

:cool: To bugzpodder
thank for your edit,but i think my solution is also strong enough.anyway,i think your solution is very good.my name is Phan BachLoc ,from Vietnam,i'm 15.5,i like to have a friend like you,cause we have the same favourite,would you like to study with me,if you have any difficult problem,tell me,i help you,and in return for me.ah,can you tell me where r u from ?and more about u?

[bugzpodder]

lol. i am a bunch of questions that i am stuck on. mind if you do some for me? :D

I can't believe that even though i spent hours on these questions, i couldn't get any except first one. btw these are all elementary math questions. calculus is NOT required.

HTML format: http://www.cms.math.ca/Competitions/MOCP/2002/prob_jul_aug.html
PDF format: http://www.cms.math.ca/Competitions/MOCP/2002/prob_jul_aug.pdf
MathML format: http://www.cms.math.ca/Competitions/MOCP/2002/prob_jul_aug.mml

btw i am from windsor ontario canada. i like computer science+math. i don't think there is anything else to me lol.

[phanbachloc]

:D ok,i've seen,and i dont think it is easy to solve,but i can also
solve some of them,now what of them that u cant solve?tell me
,so i can help u

[bugzpodder]

lol i can only do the first one. which ones can u figure out?

[phanbachloc]

:D can you tell me about the online math competition?i'd like to join for my lack of money recently?please help me as a friend

[bugzpodder]

what online math competition? i don't know any. if you mean the website i gave you, its a training program.

[phanbachloc]

are u on line now,why dont we join the chat room?

[bugzpodder]

where? gotta find it first/.

[phanbachloc]

no, i dont mean your site,i mean if you know,tell me!i can solve but i must spend about 15' to solve or more,can u wait?

[bugzpodder]

lol sure. i spent at least 3 hours on these problems and only 1. i doubt you can figure a few out in 15min though, unless you are very good (which u probably are lol).

[phanbachloc]

wat about yahoo mail or MIRC ,my nick name is esmeranda,n u?

[bugzpodder]

never tried mIRC. my yahoo id is bugz_podder (with the underscore)

[phanbachloc]

where will u join in yahoo?

[phanbachloc]

but what option will u join?

[bugzpodder]

join what??

[phanbachloc]

what categories?let's join Regionals/Canada/Canada242/Canada1,ok?

[bugzpodder]

lol i'll try if i find it

[phanbachloc]

sorry for going out without telling u,my computer got problem,i 'll talk to u later,ok?i;m really sorry

[bugzpodder]

no problem i sent an email to esm..........@yahoo.com