The second question is a little more interesting, and easier I would say.
Firstly, we know the area of a triangle is given by 1/2*h*b where h=height and b=base.
So for the triangle with vertices (0,0), (x,0), (x,f(x)), our triangle has area 1/2*x*f(x).
Substituting in the given function f(x) yields: Area = a(x) = 100x / (x^2 + 4x +25).
Differentiate with the quotient rule to yield: a'(x) = - 100(x^2 -25) / (x^2 +4x +25)^2
Set a'(x) = 0 to find points where gradient = 0, i.e. the maximum point we desire.
a'(x) = 0 => x = +5 or -5. Already we can see x = -5 will not be our solution as the question states x>0, however we will check +5 to make sure it is a maximum, that is we will differentiate a'(x) and evaluate at a''(5), with the hope it yields a negative value.
a''(x) = 200(x^3-75x-100) / (x^2 +4x +25)^3
a''(5) = -10/49. Negative value => maximum, as we desired, and expected.
Finally evaluate at a(5) = 50/7.
Conclusion: Largest area triangle of area 50/7 is achieved when x=5.