Question

Problem: Find the value of $x$ that maximizes $$f(x) = \log (-20x + 12\sqrt{x}).$$ If there is no maximum value, write "NONE".

Answer 1

The value of x is 0.09

Explanation:

Here, the given function is,

`f(x)=\log(-20x+12√(x))`

Differentiating with respect to x,

`f'(x)=(1)/(-20x+12√(x))(-20+(12)/(2√(x)))`

`=(-20+(6)/(√(x)))/(-20x+12√(x))`

`=(-10+(3)/(√(x)))/(-10x+6√(x))`

`=(-10√(x)+3)/(2√(x)(-5x+3√(x)))`

Again differentiating with respect to x,

`f''(x)=-((-10√(3)+3)(-15x+3√(3)))/(4x√(x)(-5x+3√(x)))`

For maxima or minima,

`f'(x) = 0`

`\implies (-10√(3)+3)/(2√(x)(-5x+3√(3)))=0`

`\implies x \approx 0.09`

At x = 0.09,

f''(x) = negative,

Hence, for x = 0.09, f(x) is maximum.