Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 79

${\text{Vertices: }}\left( {0,0} \right),\left( {5,0} \right),\left( {0,10} \right)$

$$\eqalign{ & \left( {\text{a}} \right){\text{From the graph shown below }} \cr & {\text{Using the points: }}\left( {0,y} \right){\text{ and }}\left( {1,8} \right){\text{ and }}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & m = \frac{{8 - y}}{{1 - 0}} = 8 - y \cr & {\text{Using the points: }}\left( {x,0} \right){\text{ and }}\left( {1,8} \right){\text{ and }}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & m = \frac{{8 - 0}}{{1 - x}} = \frac{8}{{1 - x}} \cr & {\text{Let }}m = m \cr & 8 - y = \frac{8}{{1 - x}} \cr & {\text{Solving for }}y \cr & y = 8 + \frac{8}{{x - 1}} \cr & {\text{Using the pythagorean theorem to calculate the hypotenuse}} \cr & h = \sqrt {{x^2} + {y^2}} \cr & h = \sqrt {{x^2} + {{\left( {8 + \frac{8}{{x - 1}}} \right)}^2}} ,{\text{ domain }}x > 1 \cr & {\text{The expression is minimum when the radicand is minimum}} \cr & {\text{,then let }}H{\text{ the radicand}} \cr & H = {x^2} + {\left( {8 + \frac{8}{{x - 1}}} \right)^2} \cr & {\text{Differentiate}} \cr & \frac{{dH}}{{dx}} = 2x + 2\left( {8 + \frac{8}{{x - 1}}} \right)\left( { - \frac{8}{{{{\left( {x - 1} \right)}^2}}}} \right) \cr & \frac{{dH}}{{dx}} = 2x - \frac{{16}}{{{{\left( {x - 1} \right)}^2}}}\left( {8 + \frac{8}{{x - 1}}} \right) \cr & \frac{{dH}}{{dx}} = 2x - \frac{{16}}{{{{\left( {x - 1} \right)}^2}}}\left( {\frac{{8x}}{{x - 1}}} \right) \cr & \frac{{dH}}{{dx}} = 2x - \frac{{128x}}{{{{\left( {x - 1} \right)}^3}}} \cr & {\text{Let }}\frac{{dH}}{{dx}} = 0,{\text{ then}} \cr & 2x - \frac{{128x}}{{{{\left( {x - 1} \right)}^3}}} = 0 \cr & x - \frac{{64x}}{{{{\left( {x - 1} \right)}^3}}} = 0 \cr & x{\left( {x - 1} \right)^3} - 64x = 0 \cr & x\left[ {{{\left( {x - 1} \right)}^3} - 64} \right] = 0 \cr & x = 0,{\text{ }}{\left( {x - 1} \right)^3} = 64 \to x = 5 \cr & {\text{Taking }}x = 5 \cr & {\text{Calculate }}y \cr & y = 8 + \frac{8}{{x - 1}} \to y = 8 + \frac{8}{{5 - 1}} = 10 \cr & {\text{Therefore}}{\text{, the vertices of the triangle are:}} \cr & {\text{Vertices: }}\left( {0,0} \right),\left( {5,0} \right),\left( {0,10} \right) \cr} $$