# Max value of half circle

I have the function $F(x) = \sqrt{27x - x^2}$. I need help determining the maximum value. In previous examples of quadratic functions, I have been completing the square to get the equation into the form of $a(x - h)^2 + k$ where I could then get the vertex, axis of symmetry, etc. Does the same pattern hold for a different looking equation?

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The square root function is an increasing function; therefore the maximum value of $\sqrt{g(x)}$ occurs at the same value of $x$ for which the maximum value of $g(x)$ occurs (assuming $g$ takes positive values). $$-x^2+27x = -(x^2-27x) = -(x^2 -27x + \text{something}) + \text{the same something}$$ etc.

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Clearly it does. Note that $\sqrt{f}$ is a monotone function of $f$ whenever it is defined. So you should make two steps:
1. find where $\sqrt{27x-x^2}$ is defined (say, it will be an interval $[a,b]$).
2. find the maximum of $27x-x^2$ on this interval $[a,b]$.
Second solution is geometrical: there you can find the center of this circle $(x_0,y_0=0)$ and its radius $r$. Coordinate $x_0$ of a center is an argument of maximum, radius $r = F(x_0)$ is the maximum.