Question

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

Answer 1

Is it x/2 , x/2 ?

Answer 2

Yes

Answer 3

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.

Answer 4

square rule helps

Answer 5

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

Answer 6

a+b=x
a*b=max

use calculus

Answer 7

i wouldn't

Answer 8

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

Answer 9

But it can also be the smallest

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Yeah, that's a good idea.

Answer 15

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 : \(ab < \cfrac{x^2}{4} \)

Answer 16

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

Answer 17

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

Answer 18

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 \(a=b\) ?

Answer 19

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

Answer 20

what if x<0

Answer 21

then i am probably wrong

Answer 22

I'm referring to the AM - GM inequality

Answer 23

actually i think i am still right

Answer 24

you are

Answer 25

happens about once a day, sometimes twice

Answer 26

lol

Answer 27

lol

Answer 28

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

Answer 29

Have a look at my solution :

Answer 30

No problem then :

\(\bf { a+b = x\\
\text{Now} \space , (a-b)^2 \ge 0 \\
a^2 + b^2 - 2ab \ge 0 \\
a^2 + b^2 \ge 2ab \\
a^2 + b^2 + 2ab \ge 4ab \\
(a+b)^2 \ge 4ab \\
\cfrac{(a+b)^2}{4} \ge ab \\
\cfrac{x^2}{4} \ge ab \\
} \)

\(\textbf{Proved }\) ....

Answer 31

Looks better now. Isn't it @ParthKohli @Zarkon @satellite73 :)

Answer 32

Ah! Great one again!

Answer 33

Good to hear.