ParthKohli 3 years ago How do I prove that if the sum of two numbers is $$x$$, then $$\frac{x}{2}, \frac{x}{2}$$ maximizes the product?

1. anonymous

Is it x/2 , x/2 ?

2. ParthKohli

Yes

3. 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.

4. anonymous

square rule helps

5. ParthKohli

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

6. Zarkon

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

7. anonymous

i wouldn't

8. 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

9. ParthKohli

But it can also be the smallest

10. 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

11. anonymous

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

12. anonymous

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

13. anonymous

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

14. ParthKohli

Yeah, that's a good idea.

15. 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 : $$ab < \cfrac{x^2}{4}$$

16. anonymous

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

17. ParthKohli

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

18. 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 $$a=b$$ ?

19. anonymous

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

20. Zarkon

what if x<0

21. anonymous

then i am probably wrong

22. Zarkon

I'm referring to the AM - GM inequality

23. anonymous

actually i think i am still right

24. Zarkon

you are

25. anonymous

happens about once a day, sometimes twice

26. Zarkon

lol

27. ParthKohli

lol

28. anonymous

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

29. anonymous

Have a look at my solution :

30. anonymous

No problem then : $$\bf { a+b = x\\ \text{Now} \space , (a-b)^2 \ge 0 \\ a^2 + b^2 - 2ab \ge 0 \\ a^2 + b^2 \ge 2ab \\ a^2 + b^2 + 2ab \ge 4ab \\ (a+b)^2 \ge 4ab \\ \cfrac{(a+b)^2}{4} \ge ab \\ \cfrac{x^2}{4} \ge ab \\ }$$ $$\textbf{Proved }$$ ....

31. anonymous

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

32. ParthKohli

Ah! Great one again!

33. anonymous

Good to hear.