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

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