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

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ParthKohli
 one year ago
Best ResponseYou've already chosen the best response.0Let 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.

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1i 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

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

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

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

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

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

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

Thoughts
 one year ago
Best ResponseYou've already chosen the best response.3Let 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 : \(ab < \cfrac{x^2}{4} \)

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.1nah, 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 \(a=b\) ?

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

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

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

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

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

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

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

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

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

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