Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 220: 15

$$\left( {\frac{7}{2},\sqrt {\frac{7}{2}} } \right)$$

$$\eqalign{ & {\text{Let the function be}}f\left( x \right) = \sqrt x \cr & y = \sqrt x {\text{ }}\left( {\bf{1}} \right) \cr & {\text{The distance between two points }}\left( {{x_1},{y_1}} \right){\text{ and }}\left( {{x_2},{y_2}} \right){\text{ is given}} \cr & {\text{by : }}d = \sqrt {{{\left( {{x_2} - {x_1}} \right)}^2} + {{\left( {{y_2} - {y_1}} \right)}^2}} \cr & {\text{We know the point }}\left( {{x_1},{y_1}} \right) = \left( {4,0} \right),{\text{ and }}\left( {x,y} \right) = \left( {{x_2},{y_2}} \right) \cr & d = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y - 0} \right)}^2}} {\text{ }}\left( {\bf{2}} \right) \cr & {\text{Substitute }}\sqrt x {\text{ for }}y{\text{ into equation }}\left( {\bf{2}} \right) \cr & d = \sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {\sqrt x } \right)}^2}} \cr & d = \sqrt {{x^2} - 8x + 16 + x} \cr & d = \sqrt {{x^2} - 7x + 16} \cr & {\text{If the radicand is the smallest, then }}d{\text{ is the smallest, so we need the}} \cr & {\text{minimum value of the radicand to be }}r\left( x \right) = {x^2} - 7x + 16 \cr & {\text{Differentiate}} \cr & \frac{{dr}}{{dx}} = \frac{d}{{dx}}\left[ {{x^2} - 7x + 16} \right] \cr & \frac{{dr}}{{dx}} = 2x - 7 \cr & \frac{{dr}}{{dx}} = 0 \cr & 2x - 7 = 0 \cr & {\text{Factoring}} \cr & x = \frac{7}{2} \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}r}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {2x - 7} \right] \cr & \frac{{{d^2}r}}{{d{x^2}}} = 2 \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}r}}{{d{x^2}}}} \right|_{x = \frac{7}{2}}} = 2 > 0{\text{ Relative minimum}} \cr & {\text{Let }}x = \frac{7}{2} \cr & {\text{Calculate }}y \cr & y = \sqrt x \to y = \sqrt {\frac{7}{2}} \cr & {\text{The point }}\left( {x,y} \right){\text{ nearest to }}\left( {4,0} \right){\text{ is }}\left( {x,y} \right) = \left( {\frac{7}{2},\sqrt {\frac{7}{2}} } \right) \cr} $$