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ParthKohli
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How do I prove that if the sum of two numbers is \(x\), then \(\frac{x}{2}, \frac{x}{2}\) maximizes the product?
 one year ago
 one year ago
ParthKohli Group Title
How do I prove that if the sum of two numbers is \(x\), then \(\frac{x}{2}, \frac{x}{2}\) maximizes the product?
 one year ago
 one year ago

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Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Is it x/2 , x/2 ?
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Let the two numbers be \((x  n)(x k) = x^2  nx  kx + nk= x^2  (n+k) x + nk\). I don't know if that helps.
 one year ago

modphysnoob Group TitleBest ResponseYou've already chosen the best response.0
square rule helps
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
I just have to maximize \(x^2  (n + k)x + nk\) but have no idea how.
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
a+b=x a*b=max use calculus
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i wouldn't
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i would argue by symmetry call one \(a\) and the other \(b\) with \(a+b=x\) then since this equation is symmetric in \(a\) and \(b\), i.e. you cannot tell them apart, it must be largest in the middle
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
But it can also be the smallest
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
by which i mean the product must be largest in the middle but you can write as @Zarkon said maximize \(a(xa)\) using calculus
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
can't be the smallest if \(a=x\) or \(b=x\) you get \(ab=0\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i.e. it is smallest at the endpoint of the interval
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
or for that matter you can find the max of \(axa^2\) is at the vertex, which gives \(\frac{x}{2}\)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Yeah, that's a good idea.
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Let those numbers be a and b a + b = x You have to prove that : \(ab < \cfrac{x^2}{4} \) a + b = x \(\cfrac{a+b}{2} = \cfrac{x}{2}\) Using A.M  G.M inequality : \(\cfrac{\alpha + \beta}{2} > \sqrt{\alpha \beta}\) Therefore : \(\cfrac{a+b}{2} > \sqrt{ab} \) \(\cfrac{x}{2} > \sqrt{ab}\) \(\cfrac{x^2 }{4} > ab\) or : \(ab < \cfrac{x^2}{4} \)
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Hence Proved (if that was to prove basically :) )
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
It should be \(\le\) and \(\ge\), but I get your point. Real good :)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
nah, i like symmetry more look, you call one \(a\) and the other \(b\) where \(a+b=x\) but i come along and call the first one \(b\) and the second one \(a\) and write \(b+a=x\) it is pretty clear that there is no difference between them so how could you favour \(a\) over \(b\) somehow, so that the answer would not be \(a=b\) ?
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Symmetry is good too. But I think using AM  GM inequality is also fine and good. :)
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
what if x<0
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
then i am probably wrong
 one year ago

Zarkon Group TitleBest ResponseYou've already chosen the best response.1
I'm referring to the AM  GM inequality
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
actually i think i am still right
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
happens about once a day, sometimes twice
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Done (in a clever 5th grade way) ;)
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Have a look at my solution :
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
No problem then : \(\bf { a+b = x\\ \text{Now} \space , (ab)^2 \ge 0 \\ a^2 + b^2  2ab \ge 0 \\ a^2 + b^2 \ge 2ab \\ a^2 + b^2 + 2ab \ge 4ab \\ (a+b)^2 \ge 4ab \\ \cfrac{(a+b)^2}{4} \ge ab \\ \cfrac{x^2}{4} \ge ab \\ } \) \(\textbf{Proved }\) ....
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Looks better now. Isn't it @ParthKohli @Zarkon @satellite73 :)
 one year ago

ParthKohli Group TitleBest ResponseYou've already chosen the best response.0
Ah! Great one again!
 one year ago

Thoughts Group TitleBest ResponseYou've already chosen the best response.3
Good to hear.
 one year ago
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