Answer
$${\text{Relative maximum at the point }}\left( {2,0,5} \right)$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} \cr
& \sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} {\text{ is always positive, then there is a }} \cr
& {\text{maximum when }}{\left( {x - 2} \right)^2} - {y^2} = 0{\text{ at }}\left( {2,0} \right) \cr
& {\text{Global maximum at }}\left( {0,0,1} \right).{\text{ Then,}} \cr
& {\text{Calculate the first partial derivatives of }}f\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} } \right] \cr
& {f_x}\left( {x,y} \right) = \frac{{ - \left( {x - 2} \right)}}{{\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} }} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} } \right] \cr
& {f_y}\left( {x,y} \right) = \frac{{ - y}}{{\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} }} \cr
& {\text{Setting both first partial derivatives equal to zero, we have}} \cr
& \frac{{ - \left( {x - 2} \right)}}{{\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} }} = 0 \cr
& x = 2 \cr
& \frac{{ - y}}{{\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} }} = 0 \cr
& y = 0 \cr
& {\text{The critical point is }}\left( {2,0} \right) \cr
& {\text{Find the second partial derivatives of }}f\left( {x,y} \right){\text{ and }}{f_{xy}}\left( {x,y} \right) \cr
& {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{ - \left( {x - 2} \right)}}{{\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} }}} \right] = \frac{{{y^2} - 25}}{{{{\left( {25 - {{\left( {x - 2} \right)}^2} - {y^2}} \right)}^{3/2}}}} \cr
& {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{ - y}}{{\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} }}} \right] = \frac{{{x^2} - 4x - 21}}{{{{\left( {25 - {{\left( {x - 2} \right)}^2} - {y^2}} \right)}^{3/2}}}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{ - \left( {x - 2} \right)}}{{\sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} }}} \right] = - \frac{{\left( {x - 2} \right)y}}{{{{\left( {25 - {{\left( {x - 2} \right)}^2} - {y^2}} \right)}^{3/2}}}} \cr
& {\text{By the second partials test}} \cr
& d\left( {x,y} \right) = {f_{xx}}\left( {a,b} \right){f_{yy}}\left( {a,b} \right) - {\left[ {{f_{xy}}\left( {x,y} \right)} \right]^2} \cr
& {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( {2,0} \right) \cr
& d\left( {2,0} \right) = \left( { - \frac{1}{5}} \right)\left( { - \frac{1}{5}} \right) - {\left[ 0 \right]^2} \cr
& d\left( {2,0} \right) = \frac{1}{{25}} \cr
& d > 0,{\text{ and }}{f_{xx}}\left( {2,0} \right) = - \frac{1}{5} < 0 \cr
& {\text{Then}} \cr
& f\left( {x,y} \right){\text{ has a relative maximum at }}\left( {2,0,f\left( {2,0} \right)} \right) \cr
& f\left( {2,0} \right) = \sqrt {25 - {{\left( {x - 2} \right)}^2} - {y^2}} \cr
& f\left( {2,0} \right) = 5 \cr
& {\text{Relative maximum at the point }}\left( {2,0,5} \right) \cr} $$
