Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 220: 19

$$x = 700{\text{m and }}y = 350{\text{m}}$$

$$\eqalign{ & {\text{From the image shown below:}} \cr & {\text{Let }}L{\text{ be the length of the fencing to be minimized }} \cr & L = y + x + y \cr & L = x + 2y{\text{ }}\left( {\bf{1}} \right) \cr & {\text{We know that the pasture must contain 245,000}}{{\text{m}}^2},{\text{ then}} \cr & xy = 245,000 \cr & {\text{Solving the previous equation for }}y \cr & y = \frac{{245,000}}{x} \cr & {\text{Substitute }}\frac{{36}}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr & L = x + 2\left( {\frac{{245,000}}{x}} \right) \cr & L = x + \frac{{490,000}}{x}{\text{, The domain is }}x > 0 \cr & {\text{Differentiate}} \cr & \frac{{dA}}{{dx}} = 1 - \frac{{490,000}}{{{x^2}}} \cr & {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr & 1 - \frac{{490,000}}{{{x^2}}} = 0 \cr & \frac{{490,000}}{{{x^2}}} = 1 \cr & {x^2} = 490,000 \cr & x = 700 \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {1 - \frac{{490,000}}{{{x^2}}}} \right] \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{{980,000}}{{{x^3}}} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = 700}} = \frac{{980,000}}{{{{\left( {700} \right)}^3}}} > 0{\text{ Relative minimum}} \cr & {\text{Calculate }}y \cr & y = \frac{{245,000}}{x} \to y = \frac{{245,000}}{{700}} = 350 \cr & x = 700{\text{ and }}y = 350 \cr & {\text{Therefore, the dimensions are:}} \cr & x = 700{\text{m and }}y = 350{\text{m}} \cr} $$