Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 897: 70

Answer

$$\left( {0,0} \right){\text{ and }}\left( {\frac{{\root 3 \of {192} }}{3},\frac{1}{{12}}{{\left( {192} \right)}^{2/3}}} \right)$$

Work Step by Step

$$\eqalign{ & f\left( {x,y} \right) = 3{x^3} - 12xy + {y^3} \cr & {\text{Calculating the first partial derivatives}} \cr & {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^3} - 12xy + {y^3}} \right] \cr & {f_x}\left( {x,y} \right) = 9{x^2} - 12y \cr & {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^3} - 12xy + {y^3}} \right] \cr & {f_y}\left( {x,y} \right) = - 12x + 3{y^2} \cr & {\text{Set }}{f_x}\left( {x,y} \right) = 0{\text{ and }}{f_y}\left( {x,y} \right) = 0 \cr & {\text{ }}9{x^2} - 12y = 0{\text{ }}\left( {\bf{1}} \right) \cr & - 12x + 3{y^2} = 0{\text{ }}\left( {\bf{2}} \right) \cr & {\text{Calculate }}y{\text{ into equation }}\left( {\bf{1}} \right){\text{ }} \cr & y = \frac{3}{4}{x^2} \cr & {\text{Substitute }}\frac{3}{4}{x^2}{\text{ for }}y{\text{ into }}\left( {\bf{2}} \right) \cr & - 12x + 3{\left( {\frac{3}{4}{x^2}} \right)^2} = 0 \cr & - 12x + \frac{{27}}{{16}}{x^4} = 0 \cr & {\text{Factor}} \cr & \frac{{27}}{{16}}{x^4} - 12x = 0 \cr & x\left( {\frac{{27}}{{16}}{x^3} - 12} \right) = 0 \cr & {\text{Zero - factor property}} \cr & x = 0,{\text{ }}\frac{{27}}{{16}}{x^3} - 12 = 0 \cr & \frac{{27}}{{16}}{x^3} = 12 \cr & {x^3} = \frac{{192}}{{27}} \cr & x = \frac{{\root 3 \of {192} }}{3} \cr & {\text{Evaluate }}x = 0{\text{ and }}x = \frac{{\root 3 \of {192} }}{3}{\text{ into }}y = \frac{3}{4}{x^2} \cr & x = 0 \to y = \frac{3}{4}{\left( 0 \right)^2} = 0 \to {\text{ Point }}\left( {0,0} \right) \cr & x = \frac{{\root 3 \of {192} }}{3} \to y = \frac{3}{4}{\left( {\frac{{\root 3 \of {192} }}{3}} \right)^2} = \frac{3}{{4\left( 9 \right)}}{\left( {192} \right)^{2/3}} \cr & \to {\text{ Point }}\left( {\frac{{\root 3 \of {192} }}{3},\frac{1}{{12}}{{\left( {192} \right)}^{2/3}}} \right) \cr & {\text{We obtain the points}} \cr & \left( {0,0} \right){\text{ and }}\left( {\frac{{\root 3 \of {192} }}{3},\frac{1}{{12}}{{\left( {192} \right)}^{2/3}}} \right) \cr} $$
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