Question-100565

Question Number 100565 by bemath last updated on 27/Jun/20

Commented by bobhans last updated on 28/Jun/20

let z = f (x, y)

\frac{\partial z}{\partial x} = \frac{y (x^{2} + {4 y}^{2}) - (2 x) (xy - 4 y)}{(x^{2} + {4 y}^{2})} = 0

x^{2} y + {4 y}^{3} - {2 x}^{2} y + 8 xy = 0

{4 y}^{3} - x^{2} y + 8 xy = y ({4 y}^{2} - x^{2} + 8 x) = 0

y = 0 or {4 y}^{2} = x^{2} - 8 x

\frac{\partial z}{\partial y} = \frac{(x - 4) (x^{2} + {4 y}^{2}) - 8 y (xy - 4 y)}{{(x^{2} + {4 y}^{2})}^{2}} = 0

x^{3} + {4 xy}^{2} - {4 x}^{2} - {16 y}^{2} - {8 xy}^{2} + {32 y}^{2} = 0

x^{3} - {4 xy}^{2} - {4 x}^{2} + {16 y}^{2} = 0; x^{3} + (4 - x) {4 y}^{2} - {4 x}^{2} = 0

substitute {4 y}^{2} = x^{2} - 8 x

x^{3} + (4 - x) (x^{2} - 8 x) - {4 x}^{2} = 0

x^{3} + {4 x}^{2} - 32 x - x^{3} + {8 x}^{2} - {4 x}^{2} = 0

{8 x}^{2} - 32 x = 0; 8 x (x - 4) = 0 \Rightarrow {\begin{cases} x = 0 \land y = 0 (rejected) \\ x = 4 \Rightarrow {4 y}^{2} = - 16 \Rightarrow y = \pm 2 i \end{cases}

z (4, 2 i) = \frac{8 i - 8 i}{16 + 4 (- 4)} = \frac{0}{0} ?

Answered by MJS last updated on 27/Jun/20

easy to see that - \infty < \frac{(x - 4) y}{x^{2} + 4 y^{2}} < + \infty

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