SOLUTION: The sum of the lengths of the legs of a right triangle is 40cm. What is the largest possible area for such triangle?

Area = ·base·height

Let the area by y, one leg be x and the other leg be 40-x, then

   y = x(40-x)
   y = 40x - x²
   y = 20x - x²

Write that in the form 

   y = ax² + bx + c

   y = x² + 20x + 0

Since the coefficient of x² is negative, we know that the 
parabola opens downward and therefore the vertex will be a
maximum.

The vertex of a parabola is given by 

x-coordinate of the vertex = 

y-coordinate of the vertex = what you get when you substitute
                             the x-coordinate in the equation
                             and solve for y.

So for your problem:

x-coordinate of the vertex =  =  = 20

y-coordinate of the vertex = y = (20)² + 20(20) + 0 = 200

So the maximum area that such a right triangle can have is 200 
and that will be when x = 20, which means that the other leg 40-x
will also be 40-20 = 20.  This triangle will be an isosceles right
triangle.  

Answer: Maximum area = 200cm² when both legs are 20cm each.

Edwin