Chapter 3 - Applications of Differentiation - 3.7 Exercises - Page 220: 5

$x = 21{\text{ and }}y = 7$

$$\eqalign{ & {\text{Let }}x{\text{ and }}y{\text{ the numbers}}{\text{,}} \cr & {\text{We have that:}} \cr & {\text{The product is 147}} \to {\text{ }}xy = 147{\text{ }}\left( {\bf{1}} \right) \cr & {\text{The sum of the first number plus three times the }} \cr & {\text{second number is a minimum}}{\text{, then}} \cr & \to S = x + 3y \cr & {\text{In this case}},{\text{ the objective function is the area }}S = x + 3y \cr & S = x + 3y{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Solve the equation }}\left( {\bf{1}} \right){\text{ for }}y \cr & y = \frac{{147}}{x} \cr & {\text{Substitute }}\frac{{147}}{x}{\text{ into the equation }}\left( {\bf{2}} \right) \cr & S = x + 3\left( {\frac{{147}}{x}} \right) \cr & S = x + \frac{{441}}{x} \cr & {\text{Differentiate}} \cr & \frac{{dS}}{{dx}} = 1 - \frac{{441}}{{{x^2}}} \cr & {\text{Find the critical points by solving }}\frac{{dS}}{{dx}} = 0 \cr & 1 - \frac{{441}}{{{x^2}}} = 0 \cr & {x^2} = 441 \cr & x = \pm 21 \cr & {\text{Calculate the second derivative}} \cr & \frac{{{d^2}S}}{{d{x^2}}} = \frac{{882}}{{{x^3}}} \cr & {\text{By the second derivative test:}} \cr & {\left. {\frac{{{d^2}S}}{{d{x^2}}}} \right|_{x = - 21}} = \frac{{882}}{{{{\left( { - 21} \right)}^3}}} = - \frac{2}{{21}} < 0{\text{ Relative maximum}} \cr & {\left. {\frac{{{d^2}S}}{{d{x^2}}}} \right|_{x = 21}} = \frac{{882}}{{{{\left( {21} \right)}^3}}} = \frac{2}{{21}} > 0{\text{ Relative minimum}} \cr & {\text{Taking }}x = 21 \cr & y = \frac{{147}}{x} \to y = \frac{{147}}{{21}} = 7 \cr & {\text{Therefore}}{\text{, the numbers are: }}x = 21{\text{ and }}y = 7 \cr} $$