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ParthKohli Group Title

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
  • one year ago

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  1. Thoughts Group Title
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    Is it x/2 , x/2 ?

    • one year ago
  2. ParthKohli Group Title
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    Yes

    • one year ago
  3. ParthKohli Group Title
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    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.

    • one year ago
  4. modphysnoob Group Title
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    square rule helps

    • one year ago
  5. ParthKohli Group Title
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    I just have to maximize \(x^2 - (n + k)x + nk\) but have no idea how.

    • one year ago
  6. Zarkon Group Title
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    a+b=x a*b=max use calculus

    • one year ago
  7. satellite73 Group Title
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    i wouldn't

    • one year ago
  8. satellite73 Group Title
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    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

    • one year ago
  9. ParthKohli Group Title
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    But it can also be the smallest

    • one year ago
  10. satellite73 Group Title
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    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

    • one year ago
  11. satellite73 Group Title
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    can't be the smallest if \(a=x\) or \(b=x\) you get \(ab=0\)

    • one year ago
  12. satellite73 Group Title
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    i.e. it is smallest at the endpoint of the interval

    • one year ago
  13. satellite73 Group Title
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    or for that matter you can find the max of \(ax-a^2\) is at the vertex, which gives \(\frac{x}{2}\)

    • one year ago
  14. ParthKohli Group Title
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    Yeah, that's a good idea.

    • one year ago
  15. Thoughts Group Title
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    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 : \(ab < \cfrac{x^2}{4} \)

    • one year ago
  16. Thoughts Group Title
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    Hence Proved (if that was to prove basically :) )

    • one year ago
  17. ParthKohli Group Title
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    It should be \(\le\) and \(\ge\), but I get your point. Real good :-)

    • one year ago
  18. satellite73 Group Title
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    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 \(a=b\) ?

    • one year ago
  19. Thoughts Group Title
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    Symmetry is good too. But I think using AM - GM inequality is also fine and good. :)

    • one year ago
  20. Zarkon Group Title
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    what if x<0

    • one year ago
  21. satellite73 Group Title
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    then i am probably wrong

    • one year ago
  22. Zarkon Group Title
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    I'm referring to the AM - GM inequality

    • one year ago
  23. satellite73 Group Title
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    actually i think i am still right

    • one year ago
  24. Zarkon Group Title
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    you are

    • one year ago
  25. satellite73 Group Title
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    happens about once a day, sometimes twice

    • one year ago
  26. Zarkon Group Title
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    lol

    • one year ago
  27. ParthKohli Group Title
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    lol

    • one year ago
  28. Thoughts Group Title
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    Done (in a clever 5th grade way) ;)

    • one year ago
  29. Thoughts Group Title
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    Have a look at my solution :

    • one year ago
  30. Thoughts Group Title
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    No problem then : \(\bf { a+b = x\\ \text{Now} \space , (a-b)^2 \ge 0 \\ a^2 + b^2 - 2ab \ge 0 \\ a^2 + b^2 \ge 2ab \\ a^2 + b^2 + 2ab \ge 4ab \\ (a+b)^2 \ge 4ab \\ \cfrac{(a+b)^2}{4} \ge ab \\ \cfrac{x^2}{4} \ge ab \\ } \) \(\textbf{Proved }\) ....

    • one year ago
  31. Thoughts Group Title
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    Looks better now. Isn't it @ParthKohli @Zarkon @satellite73 :)

    • one year ago
  32. ParthKohli Group Title
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    Ah! Great one again!

    • one year ago
  33. Thoughts Group Title
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    Good to hear.

    • one year ago
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