Solução A primeira etapa é encontrar todas as derivadas parciais de primeira ordem:
∂ ∂ x ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 4 x ( y − 1 ) \frac{\partial}{\partial x} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 x \left(y - 1\right) ∂ x ∂ ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 4 x ( y − 1 ) calculadora de derivada parcial ).
∂ ∂ y ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 2 x 2 + 3 y 2 − 4 y \frac{\partial}{\partial y} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 2 x^{2} + 3 y^{2} - 4 y ∂ y ∂ ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 2 x 2 + 3 y 2 − 4 y calculadora de derivada parcial ).
Em seguida, resolva o sistema { ∂ f ∂ x = 0 ∂ f ∂ y = 0 \begin{cases} \frac{\partial f}{\partial x} = 0 \\ \frac{\partial f}{\partial y} = 0 \end{cases} { ∂ x ∂ f = 0 ∂ y ∂ f = 0 { 4 x ( y − 1 ) = 0 2 x 2 + 3 y 2 − 4 y = 0 \begin{cases} 4 x \left(y - 1\right) = 0 \\ 2 x^{2} + 3 y^{2} - 4 y = 0 \end{cases} { 4 x ( y − 1 ) = 0 2 x 2 + 3 y 2 − 4 y = 0
O sistema tem as seguintes soluções reais: ( x , y ) = ( 0 , 0 ) \left(x, y\right) = \left(0, 0\right) ( x , y ) = ( 0 , 0 ) ( x , y ) = ( 0 , 4 3 ) \left(x, y\right) = \left(0, \frac{4}{3}\right) ( x , y ) = ( 0 , 3 4 ) ( x , y ) = ( − 2 2 , 1 ) \left(x, y\right) = \left(- \frac{\sqrt{2}}{2}, 1\right) ( x , y ) = ( − 2 2 , 1 ) ( x , y ) = ( 2 2 , 1 ) \left(x, y\right) = \left(\frac{\sqrt{2}}{2}, 1\right) ( x , y ) = ( 2 2 , 1 )
Agora, vamos tentar classificá-los.
Encontre todas as derivadas parciais de segunda ordem:
∂ 2 ∂ x 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 4 y − 4 \frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 y - 4 ∂ x 2 ∂ 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 4 y − 4 calculadora de derivada parcial ).
∂ 2 ∂ y ∂ x ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 4 x \frac{\partial^{2}}{\partial y\partial x} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 4 x ∂ y ∂ x ∂ 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 4 x calculadora de derivada parcial ).
∂ 2 ∂ y 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 6 y − 4 \frac{\partial^{2}}{\partial y^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right) = 6 y - 4 ∂ y 2 ∂ 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) = 6 y − 4 calculadora de derivada parcial ).
Defina a expressão D = ∂ 2 f ∂ x 2 ∂ 2 f ∂ y 2 − ( ∂ 2 f ∂ y ∂ x ) 2 = − 16 x 2 + 24 y 2 − 40 y + 16. D = \frac{\partial ^{2}f}{\partial x^{2}} \frac{\partial ^{2}f}{\partial y^{2}} - \left(\frac{\partial ^{2}f}{\partial y\partial x}\right)^{2} = - 16 x^{2} + 24 y^{2} - 40 y + 16. D = ∂ x 2 ∂ 2 f ∂ y 2 ∂ 2 f − ( ∂ y ∂ x ∂ 2 f ) 2 = − 16 x 2 + 24 y 2 − 40 y + 16.
Como D ( 0 , 0 ) = 16 D{\left(0,0 \right)} = 16 D ( 0 , 0 ) = 16 0 0 0 ∂ 2 ∂ x 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) ∣ ( ( x , y ) = ( 0 , 0 ) ) = − 4 \frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)|_{\left(\left(x, y\right) = \left(0, 0\right)\right)} = -4 ∂ x 2 ∂ 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) ∣ ( ( x , y ) = ( 0 , 0 ) ) = − 4 0 0 0 ( 0 , 0 ) \left(0, 0\right) ( 0 , 0 )
Como D ( 0 , 4 3 ) = 16 3 D{\left(0,\frac{4}{3} \right)} = \frac{16}{3} D ( 0 , 3 4 ) = 3 16 0 0 0 ∂ 2 ∂ x 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) ∣ ( ( x , y ) = ( 0 , 4 3 ) ) = 4 3 \frac{\partial^{2}}{\partial x^{2}} \left(2 x^{2} y - 2 x^{2} + y^{3} - 2 y^{2} + 2\right)|_{\left(\left(x, y\right) = \left(0, \frac{4}{3}\right)\right)} = \frac{4}{3} ∂ x 2 ∂ 2 ( 2 x 2 y − 2 x 2 + y 3 − 2 y 2 + 2 ) ∣ ( ( x , y ) = ( 0 , 3 4 ) ) = 3 4 0 0 0 ( 0 , 4 3 ) \left(0, \frac{4}{3}\right) ( 0 , 3 4 )
Como D ( − 2 2 , 1 ) = − 8 D{\left(- \frac{\sqrt{2}}{2},1 \right)} = -8 D ( − 2 2 , 1 ) = − 8 0 0 0 ( − 2 2 , 1 ) \left(- \frac{\sqrt{2}}{2}, 1\right) ( − 2 2 , 1 )
Como D ( 2 2 , 1 ) = − 8 D{\left(\frac{\sqrt{2}}{2},1 \right)} = -8 D ( 2 2 , 1 ) = − 8 0 0 0 ( 2 2 , 1 ) \left(\frac{\sqrt{2}}{2}, 1\right) ( 2 2 , 1 )
Resposta Máximo relativo ( x , y ) = ( 0 , 0 ) \left(x, y\right) = \left(0, 0\right) ( x , y ) = ( 0 , 0 ) A , f ( 0 , 0 ) = 2 f{\left(0,0 \right)} = 2 f ( 0 , 0 ) = 2 A
Mínimos relativos ( x , y ) = ( 0 , 4 3 ) ≈ ( 0 , 1.333333333333333 ) \left(x, y\right) = \left(0, \frac{4}{3}\right)\approx \left(0, 1.333333333333333\right) ( x , y ) = ( 0 , 3 4 ) ≈ ( 0 , 1.333333333333333 ) A , f ( 0 , 4 3 ) = 22 27 ≈ 0.814814814814815 f{\left(0,\frac{4}{3} \right)} = \frac{22}{27}\approx 0.814814814814815 f ( 0 , 3 4 ) = 27 22 ≈ 0.814814814814815 A
Pontos de sela ( x , y ) = ( − 2 2 , 1 ) ≈ ( − 0.707106781186548 , 1 ) \left(x, y\right) = \left(- \frac{\sqrt{2}}{2}, 1\right)\approx \left(-0.707106781186548, 1\right) ( x , y ) = ( − 2 2 , 1 ) ≈ ( − 0.707106781186548 , 1 ) A , f ( − 2 2 , 1 ) = 1 f{\left(- \frac{\sqrt{2}}{2},1 \right)} = 1 f ( − 2 2 , 1 ) = 1 A
( x , y ) = ( 2 2 , 1 ) ≈ ( 0.707106781186548 , 1 ) \left(x, y\right) = \left(\frac{\sqrt{2}}{2}, 1\right)\approx \left(0.707106781186548, 1\right) ( x , y ) = ( 2 2 , 1 ) ≈ ( 0.707106781186548 , 1 ) A , f ( 2 2 , 1 ) = 1 f{\left(\frac{\sqrt{2}}{2},1 \right)} = 1 f ( 2 2 , 1 ) = 1 A
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