Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 220: 18

$${\text{The dimensions are: }}9{\text{ by }}9$$

$$\eqalign{ & {\text{From the image shown below:}} \cr & {\text{Let }}A{\text{ be the area to be minimized }} \cr & A = \left( {y + 3} \right)\left( {x + 3} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{We know that the printed area inside the margin is }} \cr & xy = 36 \cr & {\text{Solving the previous equation for }}y \cr & y = \frac{{36}}{x} \cr & {\text{Substitute }}\frac{{36}}{x}{\text{ for }}y{\text{ into the equation }}\left( {\bf{1}} \right) \cr & A = \left( {\frac{{36}}{x} + 3} \right)\left( {x + 3} \right),{\text{ The domain is }}x > 0 \cr & A = 36 + \frac{{108}}{x} + 3x + 9 \cr & A = 3x + \frac{{108}}{x} + 49 \cr & {\text{Differentiate}} \cr & \frac{{dA}}{{dx}} = 3 - \frac{{108}}{{{x^2}}} \cr & {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr & 3 - \frac{{108}}{{{x^2}}} = 0 \cr & \frac{{108}}{{{x^2}}} = 3 \cr & {x^2} = 36 \cr & x = 6 \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ { - \frac{{60}}{{{x^2}}} + 2} \right] \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{{216}}{{{x^3}}} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = 6}} = \frac{{216}}{{{{\left( 6 \right)}^3}}} > 0{\text{ Relative minimum}} \cr & {\text{Calculate }}y \cr & y = \frac{{36}}{x} \to y = \frac{{36}}{6} = 6 \cr & x = 6{\text{ and }}y = 6 \cr & {\text{Therefore, the dimensions are:}} \cr & x + 3 = 9 \cr & y + 3 = 9 \cr} $$