Given x,y∈R^+ and ((x/5)+(y/3))((5/x)+(3/y))=139. If maximum and minimum of ((x+y)/( (√(xy)) )) is M and n respectively, then what the value of 3M−4n.


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Question Number 158259 by cortano last updated on 01/Nov/21

$G i v e n x, y \in R^{+} a n d (\frac{x}{5} + \frac{y}{3}) (\frac{5}{x} + \frac{3}{y}) = 139 .$ $I f m a x i m u m a n d m i n i m u m$ $o f \frac{x + y}{\sqrt{x y}} i s M a n d n r e s p e c t i v e l y,$ $t h e n w h a t t h e v a l u e o f 3 M - 4 n .$
Commented by tounghoungko last updated on 01/Nov/21
$⇔((x/5)+(y/3))((5/x)+(3/y))=139 ⇔ ((3x)/(5y))+((5y)/(3x))=137 → { (((x/y)=a)),(((y/x)=(1/a))) :} ⇔ ((3a)/5)+(5/(3a))=137 ...(i) ⇒((x+y)/( (√(xy)))) = (√(x/y))+(√(y/x)) = (√a)+(1/( (√a))) ⇒from (i) ⇒9a^2 −2055a+25=0 by Vietha′s rule { ((a_1 +a_2 =((2055)/9))),((a_1 ×a_2 =((25)/9))) :} let { ((max=M=(√a_1 )+(1/( (√a_1 ))))),((min=n=(√a_2 )+(1/( (√a_2 ))) )) :} (3M−4n)^2 =(3(√a_1 )+(3/( (√a_1 )))−4(√a_2 )−(4/( (√a_2 ))))^2 =(((3a_1 +3)/( (√a_1 )))−4(((a_2 +1)/( (√a_2 )))))^2 =(((3a_1 (√a_2 )+3(√a_2 )−4a_2 (√a_1 )−4(√a_1 ))/( (√(a_1 a_2 )))))^2 =((((√(a_1 a_2 ))(3(√a_1 )−4(√a_2 ))+3(√a_2 )−4(√a_1 ))^2 )/((25)/9)) =(9/(25))(a_1 a_2 (9a_1 +16a_2 −12(√(a_1 a_2 )))+9a_2 +16a_1 −12(√(a_1 a_2 ))+2(√(a_1 a_2 ))(9(√(a_1 a_2 ))−12a_1 −12a_2 +16(√(a_1 a_2 )))$
$\Leftrightarrow (\frac{x}{5} + \frac{y}{3}) (\frac{5}{x} + \frac{3}{y}) = 139$ $\Leftrightarrow \frac{3 x}{5 y} + \frac{5 y}{3 x} = 137 \to {\begin{cases} \frac{x}{y} = a \\ \frac{y}{x} = \frac{1}{a} \end{cases}$ $\Leftrightarrow \frac{3 a}{5} + \frac{5}{3 a} = 137 . . . (i)$ $\Rightarrow \frac{x + y}{\sqrt{x y}} = \sqrt{\frac{x}{y}} + \sqrt{\frac{y}{x}} = \sqrt{a} + \frac{1}{\sqrt{a}}$ $\Rightarrow f r o m (i) \Rightarrow 9 a^{2} - 2055 a + 25 = 0$ $b y {V i e t h a}^{'} s r u l e {\begin{cases} a_{1} + a_{2} = \frac{2055}{9} \\ a_{1} \times a_{2} = \frac{25}{9} \end{cases}$ $l e t {\begin{cases} m a x = M = \sqrt{a_{1}} + \frac{1}{\sqrt{a_{1}}} \\ m i n = n = \sqrt{a_{2}} + \frac{1}{\sqrt{a_{2}}} \end{cases}$ ${(3 M - 4 n)}^{2} = {(3 \sqrt{a_{1}} + \frac{3}{\sqrt{a_{1}}} - 4 \sqrt{a_{2}} - \frac{4}{\sqrt{a_{2}}})}^{2}$ $= {(\frac{3 a_{1} + 3}{\sqrt{a_{1}}} - 4 (\frac{a_{2} + 1}{\sqrt{a_{2}}}))}^{2}$ $= {(\frac{3 a_{1} \sqrt{a_{2}} + 3 \sqrt{a_{2}} - 4 a_{2} \sqrt{a_{1}} - 4 \sqrt{a_{1}}}{\sqrt{a_{1} a_{2}}})}^{2}$ $= \frac{{(\sqrt{a_{1} a_{2}} (3 \sqrt{a_{1}} - 4 \sqrt{a_{2}}) + 3 \sqrt{a_{2}} - 4 \sqrt{a_{1}})}^{2}}{\frac{25}{9}}$ $= \frac{9}{25} (a_{1} a_{2} (9 a_{1} + 16 a_{2} - 12 \sqrt{a_{1} a_{2}}) + 9 a_{2} + 16 a_{1} - 12 \sqrt{a_{1} a_{2}} + 2 \sqrt{a_{1} a_{2}} (9 \sqrt{a_{1} a_{2}} - 12 a_{1} - 12 a_{2} + 16 \sqrt{a_{1} a_{2}})$
	Answered by mr W last updated on 01/Nov/21

	$(\frac{x}{5} + \frac{y}{3}) (\frac{5}{x} + \frac{3}{y}) = 139$ $\frac{5}{3} (\frac{y}{x}) + \frac{3}{5} (\frac{x}{y}) + 2 = 139$ $l e t t = \frac{y}{x}$ $25 t + \frac{9}{t} = 137 \times 15$ $25 t^{2} - 2055 t + 9 = 0$ $t = \frac{3 (137 \pm 3 \sqrt{2085})}{10}$ $P = {(\frac{x + y}{\sqrt{x y}})}^{2} = \frac{x}{y} + \frac{y}{x} + 2 = t + \frac{1}{t} + 2$ $P_{1} = \frac{3 (137 + 3 \sqrt{2085})}{10} + \frac{10}{3 (137 + 3 \sqrt{2085})} + 2$ $= \frac{2359 - 24 \sqrt{2085}}{15} = n^{2}$ $P_{2} = \frac{3 (137 - 3 \sqrt{2085})}{10} + \frac{10}{3 (137 - 3 \sqrt{2085})} + 2$ $= \frac{2359 + 24 \sqrt{2085}}{15} = M^{2}$ $3 M - 4 n = 3 \sqrt{\frac{2359 + 24 \sqrt{2085}}{15}} - 4 \sqrt{\frac{2359 - 24 \sqrt{2085}}{15}} \approx 8.962$
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