who wants to answer this problem .. find two real numbers whose sum is 4 and whose product is a maximum ???

Question

anonymous · Answer

2 and 2 by symmetry

anonymous · Answer

if the sum was 50. then i would get it..

anonymous · Answer

since you cannot tell the difference between the two numbers, the max is when they are equal

anonymous · Answer

Think of it as a square.  A square has the largest area of any rectangle with a similar perimeter

anonymous · Answer

if you have to write something to show your teacher, put 
$$x+y=4$$ so 
$$y=4-x$$ then find the maximum of 
$$xy=x(4-x)=4x-x^2$$ max of this parabola is at the vertex, and the vertex is 
$$-\frac{bZ}{2a}=-\frac{4}{-2}=2$$

anonymous · Answer

actually rickjbr said it better. max area is a square, (but you have to know that first)

anonymous · Answer

if the sum was 50 it would turn out like this

results: x + y = 50 and P = xy = x(50 - x) = 50x - x^2. Now, dP/dx = 50 - 2x which is 0 when x = 25. Hence the two numbers are 25 and 25 and their product = 625. 

It is easy to see that this is the maximun value of the product: 25(25) = 625. Suppose we try any other number x; then the other would be 50 - x and the product would be x(50 - x) and obviously the two numbers would be 25 + x and 25 - x. Their product would be (25 + x)(25 - x) =
25^2 - x^2 = 625 - x^2 < 625 because x^2 > 0.

but the sum is 2 so lets see if you can try it out.. 

the max are equal like what @satellite & @rick
use they advice ;)