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ParthKohli
 3 years ago
How do I prove that if the sum of two numbers is \(x\), then \(\frac{x}{2}, \frac{x}{2}\) maximizes the product?
ParthKohli
 3 years 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
 3 years 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
 3 years 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
 3 years ago
Best ResponseYou've already chosen the best response.1a+b=x a*b=max use calculus

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i 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
 3 years ago
Best ResponseYou've already chosen the best response.0But it can also be the smallest

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0by which i mean the product must be largest in the middle but you can write as @Zarkon said maximize \(a(xa)\) using calculus

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i.e. it is smallest at the endpoint of the interval

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0or for that matter you can find the max of \(axa^2\) is at the vertex, which gives \(\frac{x}{2}\)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Let 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 : \(ab < \cfrac{x^2}{4} \)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Hence Proved (if that was to prove basically :) )

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0nah, 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 \(a=b\) ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Symmetry is good too. But I think using AM  GM inequality is also fine and good. :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0then i am probably wrong

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0actually i think i am still right

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0happens about once a day, sometimes twice

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Done (in a clever 5th grade way) ;)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Have a look at my solution :

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Looks better now. Isn't it @ParthKohli @Zarkon @satellite73 :)

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