The sum of two positive real numbers is 100. Find their maximum possible product?

Question

In a sequence of eleven real numbers, only two are shown. The product of every 3 successive boxes is 120. Find the sum of all the numbers in the boxes (including the two already open).

_  _  (6)  _  _  _  _  _  _  (-4) _ 

Answer 1

If the numbers may be identical, 50 and 50 => product is 2500
If the numbers must be different, 49 and 51 => product is 2499

Answer 2

All the answers I'm seeing are very empirical and seem to be have been arrived at "by experience." Although they are right, the method isn't.

What the problem asks is to maximize x*(100-x). 100x - x2.
Maxima minima (if you can use that) tell us that the maxima is at x=50. Thus, the highest value is 2500.

Otherwise, if you are aware of parabola, you would know that the above expression is a functional representation of a (open-downwards) parabola and you can easily find that the apex exists at x=50. Thus, maximum value is, again, at 2500.

Comments

Correct:
d(100x-x^2)/dx=0
x=50

Answer 3

100=1+99
=2+98
=3+97
.
.
.
.
.
. =50+50
if we done multiplication we get maximum for 50*50
so answer is

2500