# Math Help - Differentiation!!

1. ## Differentiation!!

Question:
The sum of two real numbers $x$ and $y$ is 12. Find the maximum value of their product $xy$.

2. 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

3. 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?.

4. galactus still don't know how you got $x(12-x)$. can you please explain.

5. Originally Posted by looi76
galactus still don't know how you got $x(12-x)$. can you please explain.
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).