Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 221: 23

$$\eqalign{ & \left( {\text{a}} \right)L = \sqrt {{x^2} + 4 + \frac{8}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}}} \cr & \left( {\text{b}} \right)x \approx 2.587,{\text{ }}L \approx 4.162 \cr & \left( {\text{c}} \right){\text{Vertices: }}\left( {0,0} \right),\left( {2,0} \right),\left( {0,4} \right) \cr} $$

$$\eqalign{ & \left( {\text{a}} \right){\text{From the graph shown below }} \cr & {\text{Using the points: }}\left( {0,y} \right){\text{ and }}\left( {1,2} \right){\text{ and }}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & m = \frac{{2 - y}}{{1 - 0}} = 2 - y \cr & {\text{Using the points: }}\left( {x,0} \right){\text{ and }}\left( {1,2} \right){\text{ and }}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & m = \frac{{2 - 0}}{{1 - x}} = \frac{2}{{1 - x}} \cr & {\text{Let }}m = m \cr & 2 - y = \frac{2}{{1 - x}} \cr & {\text{Solving for }}y \cr & y = 2 + \frac{2}{{x - 1}} \cr & {\text{Using the pythagorean theorem to calculate }}L \cr & L = \sqrt {{x^2} + {y^2}} \cr & L = \sqrt {{x^2} + {{\left( {2 + \frac{2}{{x - 1}}} \right)}^2}} \cr & L = \sqrt {{x^2} + 4 + \frac{8}{{x - 1}} + \frac{4}{{{{\left( {x - 1} \right)}^2}}}} ,{\text{ domain }}x > 1 \cr & \cr & \left( {\text{b}} \right){\text{Graph of }}L\left( x \right){\text{ shown below}} \cr & {\text{From the graph }}\left( {\text{b}} \right){\text{ the length }}L{\text{ is a minimum when }}x \approx 2.587 \cr & {\text{and }}L \approx 4.162 \cr & \cr & \left( {\text{c}} \right){\text{The area of the triangle is given by:}} \cr & A = \frac{1}{2}xy \cr & {\text{Substituting }}y = 2 + \frac{2}{{x - 1}} \cr & A = \frac{1}{2}x\left( {2 + \frac{2}{{x - 1}}} \right) \cr & A = x + \frac{x}{{x - 1}},{\text{ domain }}x > 1 \cr & {\text{Differentiate}} \cr & \frac{{dA}}{{dx}} = 1 + \frac{{x - 1 - x}}{{{{\left( {x - 1} \right)}^2}}} \cr & \frac{{dA}}{{dx}} = 1 - \frac{1}{{{{\left( {x - 1} \right)}^2}}} \cr & {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr & 1 - \frac{1}{{{{\left( {x - 1} \right)}^2}}} = 0 \cr & {\left( {x - 1} \right)^2} = 1 \cr & x - 1 = \pm 1 \cr & x = 0,{\text{ }}x = 2,{\text{ the domain is }}x > 1,{\text{ then take }}x = 2 \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {1 - \frac{1}{{{{\left( {x - 1} \right)}^2}}}} \right] \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{2}{{{{\left( {x - 1} \right)}^3}}} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = 2}} = \frac{2}{{{{\left( {2 - 1} \right)}^3}}} > 0{\text{ Relative minimum}} \cr & {\text{Calculate }}y \cr & y = 2 + \frac{2}{{x - 1}} \to y = y = 2 + \frac{2}{{2 - 1}} = 4 \cr & x = 2,{\text{ }}y = 4 \cr & {\text{Therefore, the vertices of the triangle are:}} \cr & {\text{Vertices: }}\left( {0,0} \right),\left( {2,0} \right),\left( {0,4} \right) \cr} $$