- ParthKohli

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

- jamiebookeater

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- anonymous

Is it x/2 , x/2 ?

- ParthKohli

Yes

- 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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## More answers

- anonymous

square rule helps

- ParthKohli

I just have to maximize \(x^2 - (n + k)x + nk\) but have no idea how.

- Zarkon

a+b=x
a*b=max
use calculus

- anonymous

i wouldn't

- 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

But it can also be the smallest

- 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

can't be the smallest
if \(a=x\) or \(b=x\) you get \(ab=0\)

- anonymous

i.e. it is smallest at the endpoint of the interval

- anonymous

or for that matter you can find the max of \(ax-a^2\) is at the vertex, which gives \(\frac{x}{2}\)

- ParthKohli

Yeah, that's a good idea.

- 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

Hence Proved (if that was to prove basically :) )

- ParthKohli

It should be \(\le\) and \(\ge\), but I get your point. Real good :-)

- 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

Symmetry is good too. But I think using AM - GM inequality is also fine and good. :)

- Zarkon

what if x<0

- anonymous

then i am probably wrong

- Zarkon

I'm referring to the AM - GM inequality

- anonymous

actually i think i am still right

- Zarkon

you are

- anonymous

happens about once a day, sometimes twice

- Zarkon

lol

- ParthKohli

lol

- anonymous

Done (in a clever 5th grade way) ;)

- anonymous

Have a look at my solution :

- 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

- ParthKohli

Ah! Great one again!

- anonymous

Good to hear.

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