when x>=0 y>=0
12=x+y>=2sqrt(x*y) (when x=y ,then x+y=2sqrt(x*y))
so x*y<=(12/2)^2=36 ,when x=6 y=6 maximum value of x*y is 36
I will stwp through this so you can see how to apply it to similar problems.
This is a beginning max min problem. We have to maximize the product given
that x+y=12.
So, we have x+y=12 and xy must be maximum.
Solve x+y=12 for, say, y. You easily get y=12-x
Now, sub that into your product formula, xy
x(12-x)
This is what must be maximized. See, it has one variable now.
Differentiate, set to 0 and solve for x. y will follow. See?.
You have the product P = xy.
You have the condition x + y = 12. From this condition, it follows that y = 12 - x. So let y = 12 - x in P = xy.
Then you have P = x (12 - x).
Then you want to find the maximum value of P. Note that the graph of P = x (12 - x) is a parabola and the maximum value occurs at its turning point ...... (although you could also use calculus to get the coordinates of the turning point).