The triangle with maximum area is a right triangle in the first quadrant,
because its legs are the values of the intercepts.





The point on the graph where x=k is 

The slope of a line tangent to the graph at the point where x=k is 

The equation through that point with that slope is



The x-coordinate of that line is 
The y-coordinate of that line is 

So  and  are the legs of the right triangle we wish
to maximize the area of.

The area of the triangle is







Since the triangle is in the first quadrant, the only feasible value of k
for which that derivative is zero is k=8. So we substitute k=8,



That works out to be 

Edwin