Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 222: 30

$$x = \frac{{36}}{\pi }{\text{in and }}y = 36{\text{in}}$$

$$\eqalign{ & {\text{From the given image:}} \cr & {\text{Let }}V{\text{ be the volume to be maximized}} \cr & V = \pi {x^2}y{\text{ }}\left( {\bf{1}} \right) \cr & {\text{We know that the perimeter }}\left( {girth} \right){\text{ is 108 inches}},{\text{ then}} \cr & P = 2\pi x + y \cr & 2\pi x + y = 108 \cr & {\text{Solving the previous equation for }}y \cr & y = 108 - 2\pi x \cr & {\text{Substitute }}108 - 2\pi x{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr & V = {x^2}\left( {108 - 2\pi x} \right) \cr & V = 108{x^2} - 2\pi {x^3} \cr & {\text{Differentiate}} \cr & \frac{{dV}}{{dx}} = 216x - 6\pi {x^2} \cr & {\text{Let }}\frac{{dV}}{{dx}} = 0,{\text{ then}} \cr & 216x - 6\pi {x^2} = 0 \cr & 6x\left( {36 - \pi x} \right) = 0 \cr & x = 0,{\text{ }}x = \frac{{36}}{\pi } \cr & {\text{Taking }}x = \frac{{36}}{\pi } \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}V}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {216x - 6\pi {x^2}} \right] \cr & \frac{{{d^2}V}}{{d{x^2}}} = 216 - 12\pi x \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}V}}{{d{x^2}}}} \right|_{x = 18}} = 216 - 12\pi \left( {\frac{{36}}{\pi }} \right) < 0{\text{ Relative maximum at }}x = \frac{{36}}{\pi } \cr & {\text{Calculate }}y \cr & y = 108 - 2\pi x \to y = 108 - 2\pi \left( {\frac{{36}}{\pi }} \right) \cr & y = 36 \cr & {\text{Therefore, the dimensions are:}} \cr & x = \frac{{36}}{\pi }{\text{in and }}y = 36{\text{in}} \cr} $$