max value
if a+2b+c=4 then what is the maxm value of ab+bc+ca
my first thought was applying am gm with (a+b) and (b+c)
ultimately giving ab+bc+ca+b^2<=4
now, the answer is 4 which can be achieved by putting b^2=0 but thats only possible when b=0 thus giving max value of ac=4. so is it mathematically correct to say max(ab+bc+ca) = 4 even when b=0
30 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.I think so yea
Like if there aren't any other conditions
nah no other constraints given
Then it should be correct
abc are real numbers , forgot to mention
Hmmm...
but is it correct to apply am gm here since its only real numbers not positive real nos.
But should be solvable no?
Yeah, if if you are sure that condition will be true for +ve numbers only.
Yea that's the issue
there are multiple alternative solutions but weird tbh
But like in order to get a maxima you alr know the values are gonna be positive
https://www.doubtnut.com/qna/1863655 , seems logical enough
Doubtnut
Let a, b, c be real numbers such that a + 2b + c = 4. Find max(ab + bc
Let a, b, c be real numbers such that a + 2b + c = 4. Find max(ab + bc + ca).
Mathematics Stack Exchange
Maximum value of $ab+bc+ca$ given that $a+2b+c=4$
Question:
Find the maximum value of $ab+bc+ca$ from the equation, $$a+2b+c=4$$
My method:
I tried making quadratic equation(in $b$) and then putting discriminant greater than or equa...
Find the maximum value of $ab+bc+ca$ from the equation, $$a+2b+c=4$$
My method:
I tried making quadratic equation(in $b$) and then putting discriminant greater than or equa...
not necessarily
like that cant be assumed in general
Well look at it just by applying common sense
b has to be 0 cuz of it's coefficient to maximize values of a & c
i mean here yeah , but in general no
That is true. That is why the approac in your links is better.
Yea exactly
aight thats the way to go then ig
Assuming $(a-c)^{2}+4b^{2} \geq 0$
SirLancelotDuLac
this seems the best
although lengthy
how do we proceed by considering the quadratic in c instead of b at -1:12 @SirLancelotDuLac
got it silly calc erros
I mean technically it is in reverse but...
The above ss'ed answer.
reaching the solution after looking at the answer
Ya exactly.
anyways thanx. closing the thread now. can u please solve my other doubt tagging u there
+solved @SirLancelotDuLac @Naiiinaaaa
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