SOLUTION: Find the value of x that maximizes f(x) = \log (-20x + 16 \sqrt{x}

SOLUTION: Find the value of x that maximizes f(x) = \log (-20x + 16 \sqrt{x} - x).

Question 1209869: Find the value of x that maximizes
f(x) = \log (-20x + 16 \sqrt{x} - x).
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**1. Understand the Domain**
First, we need to find the domain of the function inside the logarithm. We need:
* `-20x + 16√x - x > 0`
* `-21x + 16√x > 0`
* `16√x > 21x`
Since `√x` and `x` are involved, we know that `x ≥ 0`.
Let `y = √x`, so `x = y²`. Then the inequality becomes:
* `16y > 21y²`
* `16y - 21y² > 0`
* `y(16 - 21y) > 0`
This inequality holds when `0 < y < 16/21`. Since `y = √x`, we have:
* `0 < √x < 16/21`
* `0 < x < (16/21)²`
* `0 < x < 256/441`
So the domain is `0 < x < 256/441`.
**2. Maximize the Inside of the Logarithm**
Since the logarithm is an increasing function, maximizing `f(x)` is equivalent to maximizing the expression inside the logarithm:
* `g(x) = -21x + 16√x`
Let `y = √x` again. Then `g(x) = -21y² + 16y`.
This is a quadratic function in `y`. To find its maximum, we can complete the square or find the vertex.
The vertex of a quadratic `ay² + by + c` is at `y = -b / (2a)`. In our case:
* `y = -16 / (2 * -21) = 16 / 42 = 8 / 21`
Now, substitute back `y = √x`:
* `√x = 8 / 21`
* `x = (8 / 21)² = 64 / 441`
Since `64/441` is within the domain `(0, 256/441)`, this is a valid maximum.
**3. Verify the Maximum**
To ensure this is a maximum, we can take the second derivative of `g(x)` with respect to `x`:
* `g(x) = -21x + 16x^(1/2)`
* `g'(x) = -21 + 8x^(-1/2)`
* `g''(x) = -4x^(-3/2)`
Since `g''(x)` is negative for all `x > 0`, the function `g(x)` is concave down, and the value we found is indeed a maximum.
**Conclusion**
The value of `x` that maximizes `f(x) = log(-20x + 16√x - x)` is:
* `x = 64 / 441`

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