Chapter 15 - Section 15.3 - Maxima and Minima - Exercises - Page 1117: 13

$${\text{Relative maximum at the point }}\left( { - \frac{1}{2},\frac{1}{2},\frac{3}{2}} \right)$$

$$\eqalign{ & g\left( {x,y} \right) = 1 - {x^2} - x - {y^2} + y \cr & {\text{Calculate the first partial derivatives of }}g\left( {x,y} \right) \cr & {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {1 - {x^2} - x - {y^2} + y} \right] \cr & {g_x}\left( {x,y} \right) = - 2x - 1 \cr & and \cr & {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {1 - {x^2} - x - {y^2} + y} \right] \cr & {g_y}\left( {x,y} \right) = - 2y + 1 \cr & {\text{Setting both first partial derivatives equal to zero, we have}} \cr & {g_x}\left( {x,y} \right) = 0 \cr & - 2x - 1 = 0 \cr & x = - \frac{1}{2} \cr & {g_y}\left( {x,y} \right) = 0 \cr & - 2y + 1 = 0 \cr & y = \frac{1}{2} \cr & {\text{The critical point is }}\left( { - \frac{1}{2},\frac{1}{2}} \right) \cr & {\text{Find the second partial derivatives of }}g\left( {x,y} \right){\text{ and }}{g_{xy}}\left( {x,y} \right) \cr & {g_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 2x - 1} \right] = - 2 \cr & {g_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2y + 1} \right] = - 2 \cr & {g_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 2x - 1} \right] = 0 \cr & {\text{By the second partials test}} \cr & d\left( {x,y} \right) = {g_{xx}}\left( {a,b} \right){g_{yy}}\left( {a,b} \right) - {\left[ {{g_{xy}}\left( {x,y} \right)} \right]^2} \cr & {\text{Evaluate }}d\left( {x,y} \right){\text{ at the critical point }}\left( { - \frac{1}{2},\frac{1}{2}} \right) \cr & d\left( { - \frac{1}{2},\frac{1}{2}} \right) = {g_{xx}}\left( { - \frac{1}{2},\frac{1}{2}} \right){g_{yy}}\left( { - \frac{1}{2},\frac{1}{2}} \right) - {\left[ {{g_{xy}}\left( { - \frac{1}{2},\frac{1}{2}} \right)} \right]^2} \cr & d\left( { - \frac{1}{2},\frac{1}{2}} \right) = \left( { - 2} \right)\left( { - 2} \right) - {\left[ 0 \right]^2} \cr & d\left( { - \frac{1}{2},\frac{1}{2}} \right) = 4 \cr & d > 0,{\text{ and }}{g_{xx}}\left( { - \frac{1}{2},\frac{1}{2}} \right) = - 2 < 0 \cr & {\text{then}} \cr & g\left( {x,y} \right){\text{ has a relative maximum at }}\left( { - \frac{1}{2},\frac{1}{2},f\left( { - \frac{1}{2},\frac{1}{2}} \right)} \right) \cr & g\left( { - \frac{1}{2},\frac{1}{2}} \right) = 1 - {\left( { - \frac{1}{2}} \right)^2} - \left( { - \frac{1}{2}} \right) - {\left( {\frac{1}{2}} \right)^2} + \left( { - \frac{1}{2}} \right) \cr & g\left( { - \frac{1}{2},\frac{1}{2}} \right) = \frac{3}{2} \cr & {\text{Relative maximum at the point }}\left( { - \frac{1}{2},\frac{1}{2},\frac{3}{2}} \right) \cr} $$