Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 221: 20

$$x = \frac{{15}}{2}{\text{cm and }}y = \frac{{15}}{2}{\text{cm}}$$

$$\eqalign{ & {\text{From the image shown below:}} \cr & {\text{Let }}V{\text{ be the volume to be maximized}} \cr & V = {x^2}y{\text{ }}\left( {\bf{1}} \right) \cr & {\text{We know that the solid has a surface area of 337}}{\text{.5}}{{\text{ cm}}^2},{\text{ then}} \cr & 2{x^2} + 4xy = 337.5 \cr & {\text{Solving the previous equation for }}y \cr & y = \frac{{337.5 - 2{x^2}}}{{4x}} \cr & {\text{Substitute }}\frac{{337.5 - 2{x^2}}}{{4x}}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr & V = {x^2}\left( {\frac{{337.5 - 2{x^2}}}{{4x}}} \right) \cr & V = x\left( {\frac{{337.5 - 2{x^2}}}{4}} \right) \cr & V = \frac{{337.5}}{4}x - \frac{{{x^3}}}{2}{\text{, The domain is }}x > 0 \cr & {\text{Differentiate}} \cr & \frac{{dA}}{{dx}} = \frac{{337.5}}{4} - \frac{{3{x^2}}}{2} \cr & {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr & \frac{{337.5}}{4} - \frac{{3{x^2}}}{2} = 0 \cr & \frac{{3{x^2}}}{2} = \frac{{337.5}}{4} \cr & 3{x^2} = \frac{{675}}{4} \cr & {x^2} = \frac{{225}}{4} \cr & x = \frac{{15}}{2} \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {\frac{{337.5}}{4} - \frac{{3{x^2}}}{2}} \right] \cr & \frac{{{d^2}A}}{{d{x^2}}} = - 3x \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = \frac{{15}}{2}}} = - 3\left( {\frac{{15}}{2}} \right) < 0{\text{ Relative maximum}} \cr & {\text{Calculate }}y \cr & y = \frac{{337.5 - 2{x^2}}}{{4x}} \to y = \frac{{337.5 - 2{{\left( {15/2} \right)}^2}}}{{4\left( {15/2} \right)}} = \frac{{15}}{2} \cr & x = \frac{{15}}{2}{\text{ and }}y = \frac{{15}}{2} \cr & {\text{Therefore, the dimensions are:}} \cr & x = \frac{{15}}{2}{\text{cm and }}y = \frac{{15}}{2}{\text{cm}} \cr} $$