Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 220: 11

$${\text{Width}} = {\text{length}} = 4\sqrt 2 {\text{ft}}$$

$$\eqalign{ & {\text{From the image shown below we know that the perimeter is}} \cr & {\text{given by the equation}} \cr & xy = 32{\text{ }}\left( {\bf{1}} \right) \cr & P = 2x + 2y{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solve equation }}\left( {\bf{1}} \right){\text{ for }}y \cr & y = \frac{{32}}{x} \cr & {\text{Substitute }}5 - x{\text{ into equation }}\left( {\bf{2}} \right) \cr & P = 2x + 2\left( {\frac{{32}}{x}} \right) \cr & P = 2x + \frac{{64}}{x} \cr & {\text{Differentiate}} \cr & \frac{{dP}}{{dx}} = 2 - \frac{{64}}{{{x^2}}} \cr & {\text{Find the critical points by solving }}\frac{{dA}}{{dx}} = 0 \cr & 2 - \frac{{64}}{{{x^2}}} = 0 \cr & {x^2} = \frac{{64}}{2} \cr & {x^2} = 32 \cr & x = \pm \sqrt {32} \cr & x = \pm 4\sqrt 2 \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}P}}{{d{x^2}}} = \frac{d}{{dx}}\left[ {2 - \frac{{64}}{{{x^2}}}} \right] \cr & \frac{{{d^2}P}}{{d{x^2}}} = \frac{{128}}{{{x^3}}} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}P}}{{d{x^2}}}} \right|_{x = - 4\sqrt 2 }} = \frac{{128}}{{{{\left( { - 4\sqrt 2 } \right)}^3}}} < 0{\text{ Relative maximum}} \cr & {\left. {\frac{{{d^2}P}}{{d{x^2}}}} \right|_{x = 4\sqrt 2 }} = \frac{{128}}{{{{\left( {4\sqrt 2 } \right)}^3}}} > 0{\text{ Relative minimum}} \cr & {\text{Taking }}x = 4\sqrt 2 \cr & y = \frac{{32}}{x} \to y = \frac{{32}}{{4\sqrt 2 }} = 4\sqrt 2 \cr & {\text{Therefore, the numbers are: }}x = 4\sqrt 2 {\text{ and }}y = 4\sqrt 2 \cr & {\text{The dimensions are Width}} = {\text{length}} = 4\sqrt 2 {\text{ft}} \cr} $$