Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 220: 17

$${\text{The dimensions are: }}\left( {2 + \sqrt {30} } \right){\text{ by }}\left( {2 + \sqrt {30} } \right)$$

$$\eqalign{ & {\text{From the image shown below:}} \cr & {\text{Let }}A{\text{ be the area to be minimized }} \cr & A = \left( {y + 2} \right)\left( {x + 2} \right){\text{ }}\left( {\bf{1}} \right) \cr & {\text{We know that the printed area inside the margin is }} \cr & xy = 30 \cr & {\text{Solving the previous equation for }}y \cr & y = \frac{{30}}{x} \cr & {\text{Substitute }}\frac{{30}}{x}{\text{ for }}y{\text{ in the equation }}\left( {\bf{1}} \right) \cr & A = \left( {\frac{{30}}{x} + 2} \right)\left( {x + 2} \right),{\text{ The domain is }}x > 0 \cr & A = 30 + \frac{{60}}{x} + 2x + 4 \cr & A = 34 + \frac{{60}}{x} + 2x \cr & {\text{Differentiate}} \cr & \frac{{dA}}{{dx}} = - \frac{{60}}{{{x^2}}} + 2 \cr & {\text{Let }}\frac{{dA}}{{dx}} = 0,{\text{ then}} \cr & - \frac{{60}}{{{x^2}}} + 2 = 0 \cr & \frac{{60}}{{{x^2}}} = 2 \cr & {x^2} = 30 \cr & x = \sqrt {30} \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{d}{{dx}}\left[ { - \frac{{60}}{{{x^2}}} + 2} \right] \cr & \frac{{{d^2}A}}{{d{x^2}}} = \frac{{120}}{{{x^3}}} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}A}}{{d{x^2}}}} \right|_{x = \sqrt {30} }} = \frac{{120}}{{{{\left( {\sqrt 3 } \right)}^3}}} > 0{\text{ Relative minimum}} \cr & {\text{Calculate }}y \cr & y = \frac{{30}}{x} \to y = \frac{{30}}{{\sqrt {30} }} = 10\sqrt 3 \cr & x = \sqrt {30} {\text{ and }}y = \sqrt {30} \cr & {\text{Therefore, the dimensions are:}} \cr & x + 2 = {\text{2 + }}\sqrt {30} \cr & y + 2 = {\text{2 + }}\sqrt {30} \cr} $$