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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?

Mathematics
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Is it x/2 , x/2 ?
Yes
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.

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Other answers:

square rule helps
I just have to maximize \(x^2 - (n + k)x + nk\) but have no idea how.
a+b=x a*b=max use calculus
i wouldn't
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
But it can also be the smallest
by which i mean the product must be largest in the middle but you can write as @Zarkon said maximize \(a(x-a)\) using calculus
can't be the smallest if \(a=x\) or \(b=x\) you get \(ab=0\)
i.e. it is smallest at the endpoint of the interval
or for that matter you can find the max of \(ax-a^2\) is at the vertex, which gives \(\frac{x}{2}\)
Yeah, that's a good idea.
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 : \(ab < \cfrac{x^2}{4} \)
Hence Proved (if that was to prove basically :) )
It should be \(\le\) and \(\ge\), but I get your point. Real good :-)
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 \(a=b\) ?
Symmetry is good too. But I think using AM - GM inequality is also fine and good. :)
what if x<0
then i am probably wrong
I'm referring to the AM - GM inequality
actually i think i am still right
you are
happens about once a day, sometimes twice
lol
lol
Done (in a clever 5th grade way) ;)
Have a look at my solution :
No problem then : \(\bf { a+b = x\\ \text{Now} \space , (a-b)^2 \ge 0 \\ a^2 + b^2 - 2ab \ge 0 \\ a^2 + b^2 \ge 2ab \\ a^2 + b^2 + 2ab \ge 4ab \\ (a+b)^2 \ge 4ab \\ \cfrac{(a+b)^2}{4} \ge ab \\ \cfrac{x^2}{4} \ge ab \\ } \) \(\textbf{Proved }\) ....
Looks better now. Isn't it @ParthKohli @Zarkon @satellite73 :)
Ah! Great one again!
Good to hear.

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