Chapter 3 - Applications of Differentiation - Review Exercises - Page 240: 77

$$x = 50{\text{ft and }}y = \frac{{200}}{3}{\text{ft}}$$

$$\eqalign{ & {\text{From the image shown below:}} \cr & {\text{Let }}A{\text{ be the area to be maximized}} \cr & A = 2xy{\text{ }}\left( {\bf{1}} \right) \cr & {\text{We know that the the length of the fencing is 400ft}},{\text{ then}} \cr & 4x + 3y = 400 \cr & {\text{Solving the previous equation for }}y \cr & y = \frac{{400 - 4x}}{3} \cr & {\text{Substitute }}\frac{{400 - 4x}}{3}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr & A = 2x\left( {\frac{{400 - 4x}}{3}} \right) \cr & A = \frac{{800x}}{3} - \frac{{8{x^2}}}{3}{\text{, The domain is }}0 < x < 100 \cr & {\text{Differentiate}} \cr & \frac{{dA}}{{dx}} = \frac{{800}}{3} - \frac{{16x}}{3} \cr & {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr & \frac{{800}}{3} - \frac{{16x}}{3} = 0 \cr & 16x = 800 \cr & x = 50 \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {\frac{{800}}{3} - \frac{{16x}}{3}} \right] \cr & \frac{{{d^2}A}}{{d{x^2}}} = - \frac{{16}}{3} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = 50}} = - \frac{{16}}{3} < 0{\text{ Relative maximum}} \cr & {\text{Calculate }}y \cr & y = \frac{{400 - 4x}}{3} \to y = \frac{{400 - 4\left( {50} \right)}}{3} = \frac{{200}}{3} \cr & x = 50{\text{ and }}y = \frac{{200}}{3} \cr & {\text{Therefore, the dimensions are:}} \cr & x = 50{\text{ft and }}y = \frac{{200}}{3}{\text{ft}} \cr} $$