MAIN CONCEPTS
Due to the limitation to our algebraic methods, we have to switch to a graphically method to solve this question. We thus graph out the rather complicated function by completing the square of the terms.
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In this question,  is given by two quadratic terms under square roots which is a killer to simplify and so answer to this question requires a different approach. One way is to consider completing the square for the individual quadratic terms.

and

For  to be in the domain of , we must have both

, which implies

and similarly

, which implies

 

We consider the values of  in the interval [6,8] and for these values we have

At this juncture, we switch to a graphical analysis as our algebraic methods may prove insufficient.

Graphing out  and  separately.

 

As shown in the figure, the graph of  is the semi-circle in the first quadrant with the center at (4,0) and radius 4. The graph of  is the semi-circle in the first quadrant with the centre (7,0) and radius 1.

Since our function is valid in the domain [6,8], we only need to look at that portion of the graph which greatly simplifies our analysis. In this way, it is clear that the value in [6,8] that maximizes  is  since this value maximizes  and also minimizes . hence, the maximum value of  is

This is not such an easy question. Students of calculus may be tempted to use differentiation to find the maximum value. While this is perfectly valid, it may be tedious and prone to error for students who are not proficient in the technique. It suffices to say that no problem on the AMC has a calculus solution that is easier than some non-calculus solution.