Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 80

$$h \approx 12.701{\text{ft}}$$

$$\eqalign{ & {\text{From the image shown below }} \cr & {\text{Using the points: }}\left( {0,y} \right){\text{ and }}\left( {4,5} \right){\text{ and }}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & m = \frac{{5 - y}}{{4 - 0}} = \frac{{5 - y}}{4} \cr & {\text{Using the points: }}\left( {x,0} \right){\text{ and }}\left( {4,5} \right){\text{ and }}m = \frac{{{y_2} - {y_1}}}{{{x_2} - {x_1}}} \cr & m = \frac{{5 - 0}}{{4 - x}} = \frac{5}{{4 - x}} \cr & {\text{Let }}m = m \cr & \frac{{5 - y}}{4} = \frac{5}{{4 - x}} \cr & {\text{Solving for }}y \cr & 5 - y = \frac{{20}}{{4 - x}} \cr & y = 5 - \frac{{20}}{{4 - x}} \cr & {\text{Using the pythagoream theorem to calculate the length}} \cr & {\text{of the beam}} \cr & h = \sqrt {{x^2} + {y^2}} \cr & h = \sqrt {{x^2} + {{\left( {5 - \frac{{20}}{{4 - x}}} \right)}^2}} ,{\text{ domain }}x > 4 \cr & {\text{The expression is a minimum when the radicand is a minimum,}} \cr & {\text{ so let }}L{\text{ be the radicand}} \cr & L = {x^2} + {\left( {5 - \frac{{20}}{{4 - x}}} \right)^2} \cr & {\text{Differentiate}} \cr & \frac{{dL}}{{dx}} = 2x + 2\left( {5 - \frac{{20}}{{4 - x}}} \right)\left( {\frac{{20}}{{{{\left( {4 - x} \right)}^2}}}} \right) \cr & \frac{{dL}}{{dx}} = 2x + 2\left( {\frac{{20 - 5x - 20}}{{4 - x}}} \right)\left( {\frac{{20}}{{{{\left( {4 - x} \right)}^2}}}} \right) \cr & \frac{{dL}}{{dx}} = 2x + \frac{{200x}}{{{{\left( {x - 4} \right)}^3}}} \cr & {\text{Let }}\frac{{dL}}{{dx}} = 0,{\text{ then}} \cr & 2x + \frac{{200x}}{{{{\left( {x - 4} \right)}^3}}} = 0 \cr & x + \frac{{100x}}{{{{\left( {x - 4} \right)}^3}}} = 0 \cr & x{\left( {x - 4} \right)^3} + 100x = 0 \cr & x\left[ {{{\left( {x - 4} \right)}^3} - 100} \right] = 0 \cr & x = 0,{\text{ }}{\left( {x - 4} \right)^3} - 100 = 0 \cr & {\left( {x - 4} \right)^3} = 100 \cr & x - 4 = \root 3 \of {100} \cr & x = 4 + \root 3 \of {100} {\text{ and }}x = 0{\text{ }}\left( {{\text{Domain }}x > 4} \right) \cr & {\text{Taking }}x = 4 + \root 3 \of {100} \cr & {\text{The length of the shortest beam is:}} \cr & h = \sqrt {{x^2} + {{\left( {5 - \frac{{20}}{{4 - x}}} \right)}^2}} \cr & h = \sqrt {{{\left( {4 + \root 3 \of {100} } \right)}^2} + {{\left( {5 - \frac{{20}}{{4 - 4 + \root 3 \of {100} }}} \right)}^2}} \cr & {\text{Simplifying}} \cr & h \approx 12.701{\text{ft}} \cr} $$