Given-x-y-R-x-y-0-Find-the-maximum-and-minimum-value-of-xy-4y-2-x-2-4y-2-

Question Number 132753 by liberty last updated on 16/Feb/21

Given x, y \in R, x, y \neq 0

Find the maximum and minimum

value of \frac{xy - {4 y}^{2}}{x^{2} + {4 y}^{2}}

Commented by mr W last updated on 16/Feb/21

m i n = - \frac{\sqrt{5} + 2}{4}

m a x = \frac{\sqrt{5} - 2}{4}

Commented by liberty last updated on 16/Feb/21

step ?

Answered by EDWIN88 last updated on 16/Feb/21

let z = f (x, y) = \frac{xy - {4 y}^{2}}{x^{2} + {4 y}^{2}}

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

we get x^{2} y + {4 y}^{3} - {2 x}^{2} y + {8 xy}^{2} = 0 \dots (i)

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

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

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

x^{3} - {4 xy}^{2} - 8 xy = 0 or x^{2} - {4 y}^{2} - 8 y = 0 \dots (ii)

adding (1) and (2)

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

y = - 1 \pm \frac{\sqrt{5}}{2} then critical point is

A (1, - 1 + \frac{\sqrt{5}}{2}) and B = (1, - 1 - \frac{\sqrt{5}}{2})

for A \Rightarrow z = \frac{y (x - 4 y)}{x^{2} + {4 y}^{2}} = \frac{\sqrt{5} - 2}{4} \approx 0.05901

for B \Rightarrow z = \frac{- \sqrt{5} - 2}{4} \approx - 1.05901

Commented by EDWIN88 last updated on 16/Feb/21

yes sir . thank you

Commented by liberty last updated on 16/Feb/21

waw \dots

Commented by liberty last updated on 16/Feb/21

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

z = \frac{\frac{\sqrt{5} - 2}{2} - (9 - 4 \sqrt{5})}{1 + 9 - 4 \sqrt{5}} = \frac{\sqrt{5} - 2 - 18 + 8 \sqrt{5}}{20 - 8 \sqrt{5}}

z = \frac{9 \sqrt{5} - 20}{20 - 8 \sqrt{5}} \times \frac{20 + 8 \sqrt{5}}{20 + 8 \sqrt{5}}

z = \frac{20 \sqrt{5} + 360 - 400}{400 - 320} = \frac{20 \sqrt{5} - 40}{80}

z = \frac{\sqrt{5} - 2}{4} sir

Answered by mr W last updated on 16/Feb/21

l e t k = \frac{xy - {4 y}^{2}}{x^{2} + {4 y}^{2}} = \frac{1 - 4 (\frac{y}{x})}{(\frac{x}{y}) + 4 (\frac{y}{x})} = \frac{1 - 4 t}{\frac{1}{t} + 4 t} = \frac{t - 4 t^{2}}{1 + 4 t^{2}}

w i t h t = \frac{y}{x} \in R

k + 4 {k t}^{2} = t - 4 t^{2}

4 (k + 1) t^{2} - t + k = 0

Δ = 1^{2} - 4 \times 4 (k + 1) k ⩾ 0 s u c h t h a t t \in R

k^{2} + k - \frac{1}{16} ⩽ 0

k_{1, 2} = \frac{1}{2} (- 1 \pm \frac{\sqrt{5}}{2}) = - \frac{\sqrt{5} + 2}{4}, \frac{\sqrt{5} - 2}{4}

\Rightarrow - \frac{\sqrt{5} + 2}{4} ⩽ k ⩽ \frac{\sqrt{5} - 2}{4}

\Rightarrow m i n = - \frac{\sqrt{5} + 2}{4}

\Rightarrow m a x = \frac{\sqrt{5} - 2}{4}

Commented by liberty last updated on 16/Feb/21

thank you sir

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