Inspired From Inspiration

The graph of $f\left( x \right)$ is given by Graph And the maximum value occurs at ${ x }_{ 0 }=5$ and maximum value is ${ y }_{ 0 }=4$ So our answer is ${ x }_{ 0 }+{ y }_{ 0 }=5+4=9$

Let $S_1 = \sqrt{25-(x-2)^2}, S_2= \sqrt{\frac{49}{4} - (x-\frac{3}{2})^2}$

The given function $f(x) = S_1 - S_2$ is the difference in ordinates of the points lying on the two semicircles $S_1$ and $S_2$ having the same abscissa.

Now let's collect some data about the two semicircles. The semicircle $S_2$ has centre $C_2(\frac{3}{2},0)$ , radius $R_2=\frac{7}{2}$ and is contained within semicircle $S_1$ which is centered at $C_1(2,0)$ with radius $R_1=5$ . Both lie above y-axis and the domain of the function is $x \in [-2,5]$ .

The ordinate of such a semicircle decreases as we move the along the x-axis, away from the centre of the semicircle $\ldots \{1\}$

Also, for the same change in $x$ , due to a larger radius, magnitude of increase/decrease in $S_1$ is less than that in $S_2$ $\ldots \{2\}$

From $\{1\}$ , we conclude maximum value of $f(x)$ occurs when we select an $x$ which is farther away from $C_2$ and closer to $C_1$ ,i.e., $x>\frac{3}{2}$

Also from $\{2\}$ , we notice that as we move towards the right from $x=2$ , $S_2$ decreases more rapidly than $S_1$ , thus the difference in their ordinates keep increasing in the interval $x>2$ .

The maximum value permitted for $x>2$ is $x=5$ and hence, maximum value of the function is:

$f(5) = \sqrt{25-(5-2)^2} - 0 = \sqrt{16} = 4$

Hence,

$x_0 + y_0 = 5 + 4 = \boxed{9}$

Note :- For better visualisation:

1)Sketch the graphs of $S_1$ and $S_2$ .

2)Picture the part of the line $x=k, k \in [-2,5]$ intercepted between $S_1$ and $S_2$ as $k$ varies over it's domain.

Feel free to ask for any further clarification.