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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?
 10 months ago
 10 months ago
How do I prove that if the sum of two numbers is \(x\), then \(\frac{x}{2}, \frac{x}{2}\) maximizes the product?
 10 months ago
 10 months ago

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ParthKohliBest 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.
 10 months ago

modphysnoobBest ResponseYou've already chosen the best response.0
square rule helps
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
I just have to maximize \(x^2  (n + k)x + nk\) but have no idea how.
 10 months ago

ZarkonBest ResponseYou've already chosen the best response.1
a+b=x a*b=max use calculus
 10 months ago

satellite73Best 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
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
But it can also be the smallest
 10 months ago

satellite73Best 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
 10 months ago

satellite73Best ResponseYou've already chosen the best response.1
can't be the smallest if \(a=x\) or \(b=x\) you get \(ab=0\)
 10 months ago

satellite73Best ResponseYou've already chosen the best response.1
i.e. it is smallest at the endpoint of the interval
 10 months ago

satellite73Best 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 \(\frac{x}{2}\)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Yeah, that's a good idea.
 10 months ago

ThoughtsBest 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 : \(ab < \cfrac{x^2}{4} \)
 10 months ago

ThoughtsBest ResponseYou've already chosen the best response.3
Hence Proved (if that was to prove basically :) )
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
It should be \(\le\) and \(\ge\), but I get your point. Real good :)
 10 months ago

satellite73Best 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 \(a=b\) ?
 10 months ago

ThoughtsBest ResponseYou've already chosen the best response.3
Symmetry is good too. But I think using AM  GM inequality is also fine and good. :)
 10 months ago

satellite73Best ResponseYou've already chosen the best response.1
then i am probably wrong
 10 months ago

ZarkonBest ResponseYou've already chosen the best response.1
I'm referring to the AM  GM inequality
 10 months ago

satellite73Best ResponseYou've already chosen the best response.1
actually i think i am still right
 10 months ago

satellite73Best ResponseYou've already chosen the best response.1
happens about once a day, sometimes twice
 10 months ago

ThoughtsBest ResponseYou've already chosen the best response.3
Done (in a clever 5th grade way) ;)
 10 months ago

ThoughtsBest ResponseYou've already chosen the best response.3
Have a look at my solution :
 10 months ago

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

ThoughtsBest ResponseYou've already chosen the best response.3
Looks better now. Isn't it @ParthKohli @Zarkon @satellite73 :)
 10 months ago

ParthKohliBest ResponseYou've already chosen the best response.0
Ah! Great one again!
 10 months ago
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