Find all relative extrema of the function. Use the Second-Derivative Test when applicable. (If an answer does not exist, enter DNE.)

f'(x) = 30x - 3x^2 = 0

10x - x^2 = 0

x^2 - 10x = 0

x(x - 10) = =

x = 0

x = 10

Now use second derivative test to determine maxima or minima

f"(x) = 30 - 6x

f"(0) = 30 - 6(0) = 30 which is POSITIVE

Therefore, at x = 0 there is a local MINIMUM at the point (0,0)

f"(10) = 30 - 6(10) = -30 which is NEGATIVE

Therefore, at x = 10 there is a local MAXIMUM at the point (10, 500)

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