Given-x-y-real-number-such-that-0-lt-y-x-lt-1-2-Find-minimum-value-of-2y-x-y-3x-x-2y-

Question Number 150963 by EDWIN88 last updated on 17/Aug/21

G i v e n x, y r e a l n u m b e r s u c h t h a t

0 < \frac{y}{x} < \frac{1}{2} . F i n d m i n i m u m v a l u e

o f \frac{2 y}{x - y} + \frac{3 x}{x + 2 y} .

Answered by john_santu last updated on 17/Aug/21

let \frac{y}{x} = t \Rightarrow f (t) = \frac{2 t}{1 - t} + \frac{3}{1 + 2 t}

where 0 < t < \frac{1}{2} .

taking derivative of f (t)

f^{'} (t) = \frac{2}{{(1 - t)}^{2}} - \frac{6}{{(1 + 2 t)}^{2}} = 0

\Rightarrow {(1 + 2 t)}^{2} = 3 {(1 - t)}^{2}

\Rightarrow {4 t}^{2} + 4 t + 1 = 3 - 6 t + {3 t}^{2}

\Rightarrow t^{2} + 10 t - 2 = 0

\Rightarrow {(t + 5)}^{2} - 27 = 0 \Rightarrow t = - 5 \pm 3 \sqrt{3}

f {(t)}_{\min} when t = - 5 + 3 \sqrt{3}

so \min f (t) = \frac{2 (- 5 + 3 \sqrt{3})}{1 - (- 5 + 3 \sqrt{3})} + \frac{3}{1 + 2 (- 5 + 3 \sqrt{3})}

\min f (t) = \frac{6 \sqrt{3} - 10}{6 - 3 \sqrt{3}} + \frac{1}{2 \sqrt{3} - 3}

\min f (t) = \frac{(6 \sqrt{3} - 10) (6 + 3 \sqrt{3})}{9} + \frac{2 \sqrt{3} + 3}{3}

\min f (t) = \frac{6 \sqrt{3} - 6}{9} + \frac{2 \sqrt{3} + 3}{3}

\min f (t) = \frac{2 \sqrt{3} - 2}{3} + \frac{2 \sqrt{3} + 3}{3} = \frac{4 \sqrt{3} + 1}{3}