+116 Find the value of x that maximizes
f(x) = \log (-20x + 16 \sqrt{x} - x).
+130466 f(x)=log(−20x+16√x−x)
f'x = 1 / [ -20x + 16sqrt x - x] * ln 10 = ln 10 * ( -20x + 16sqrt x - x)^-1 * [ -20 + 8 / x^(1/2) - 1 ]
Simplify and set to 0 to 0
-20 + 8 / (x)^(1/2) - 1 = 0
-21 + 8 / x^(1/2) = 0
8/ x^(1/2) = 21
x^(1/2) = 8/21 square both sides
x = (64 / 441)
