Take the derivative:
g’(x) = 12x^3 - 24x^2
Set equal to zero and solve:
0 = 12x^3 - 24x^2
0 = 12x^2 (x - 2)
x = 0 or x = 2
Plug back into original
g(0) = 3(0^4) - 8(0^3)
g(0) = 0 - 0
g(0) = 0
g(2) = 3(2^2) - 8(2^3)
g(2) = 3(4) - 8(8)
g(2) = 12 - 64
g(2) = -52
There is an absolute max at (0,0) or when x = 0
First, you must understand the question. what % of 60 is 15, so I would write an equation to express this.
15 = 60x
From there, solve for x, so Divide by 60 on each side. The answer is 1/4, or 25%
