Given-z-xy-4y-2-x-2-4y-2-x-y-0-find-minimum-and-maximum-value-of-z-

Question Number 96306 by bemath last updated on 31/May/20

Given z = \frac{xy - {4 y}^{2}}{x^{2} + {4 y}^{2}}, x, y \neq 0

find minimum and maximum

value of z

Commented by john santu last updated on 31/May/20

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

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

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

y (4 y^{2} + 8 x y - x^{2}) = 0 \Rightarrow 4 y^{2} + 8 x y - x^{2} = 0

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

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

x (x^{2} - 4 y^{2} - 8 y) = 0 \Rightarrow x^{2} - 4 y^{2} - 8 y = 0

a d d i n g (1) a n d (2)

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

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

- - - - - - - - - +

8 x y - 8 y = 0 \Rightarrow 8 y (x - 1) = 0

x = 1 \Rightarrow 4 y^{2} + 8 y - 1 = 0

y^{2} + 2 y - \frac{1}{4} = 0

{(y + 1)}^{2} = \frac{5}{4}; y = - 1 \pm \frac{\sqrt{5}}{2}

we get critical point are

(1, - 1 + \frac{\sqrt{5}}{2}) & (1, - 1 - \frac{\sqrt{5}}{2})

z (1, - 1 + \frac{\sqrt{5}}{2}) = \frac{\frac{- 2 + \sqrt{5}}{2} - 4 (\frac{9}{4} - \sqrt{5})}{1 + \frac{9}{4} - \sqrt{5}}

= \frac{- 4 + 2 \sqrt{5} - 36 + 16 \sqrt{5}}{13 - 4 \sqrt{5}} = \frac{18 \sqrt{5} - 40}{13 - 4 \sqrt{5}} \approx 0.061

z (1, - 1 - \frac{\sqrt{5}}{2}) = \frac{\frac{- 2 - \sqrt{5}}{2} - 4 (\frac{9}{4} + \sqrt{5})}{1 + \frac{9}{4} + \sqrt{5}} = \frac{- 4 - 2 \sqrt{5} - 36 - 16 \sqrt{5}}{13 + 4 \sqrt{5}}

= \frac{- 18 \sqrt{5} - 40}{13 + 4 \sqrt{5}} \approx - 3.6569