If the line 4x-9y=0 is tangent in the first quadrant to the graph of y =1/3x^3+c what is c?
slope of the tangent = dy/dx for y = (1/3)x^3 + c
= x^2
but from 4x-9y = 0 we know that slope must be 4/9
x^2 = 4/9
x = 2/3
then 4(2/3) - 9y = 0
y = 8/27
so we know the point of contact is (2/3, 8/27) which must lie on both the cubic function and the straight line.
sub into the cubic
8/27 = (1/3)(8/27) + c
c = 16/81
confirmation:
www.wolframalpha.com/input/?i=graph+4x+%3D+9y%2C+y+%3D+%281%2F3%29x%5E3+%2B+16%2F81