Find-the-minimum-of-12-x-18-y-xy-for-all-positive-number-x-amp-y-

Question Number 141135 by bobhans last updated on 16/May/21

F i n d t h e m i n i m u m o f

\frac{12}{x} + \frac{18}{y} + x y f o r a l l

p o s i t i v e n u m b e r x & y .

Answered by mitica last updated on 16/May/21

\frac{12}{x} + \frac{18}{y} + x y ⩾ 3 \sqrt[3]{\frac{12}{x} \cdot \frac{18}{y} \cdot x y} = 18

= f o r \frac{12}{x} = \frac{18}{y} = x y \Leftrightarrow x = 2, y = 3

\Rightarrow m i n = 18

Commented by mitica last updated on 16/May/21

x = 2, y = 3

Commented by iloveisrael last updated on 16/May/21

x = \sqrt[3]{\frac{4}{3}}; y = \sqrt[3]{\frac{9}{2}}

\Rightarrow \frac{12}{\sqrt[3]{\frac{4}{3}}} + \frac{18}{\sqrt[3]{\frac{9}{2}}} + \sqrt[3]{\frac{36}{6}} = 23.62

h o w y o u c l a i m 23.62 = 18 ? ?

Answered by iloveisrael last updated on 16/May/21

W e k n o w t h a t t h e t h r e e t e r m s

i n t h e s u m h a v e f i x e d p r o d u c t

12 \times 18 = 216 . T h e s u m i s t h e r e f o r e

a m i n i m u m i f w e c a n f i n d x & y

s a t i s f y \frac{12}{x} = \frac{18}{y} = x y . S i n c e t h e

p r o d u c t i s 6^{3} e a c h o f \frac{12}{x}, \frac{18}{y} & x y

m u s t e q u a l 6, h e n c e t h e r e q u i r e d

m i n i m u m i s 6 + 6 + 6 = 18,

o c c u r r i n g i f f x = 2 & y = 3 .

Answered by EDWIN88 last updated on 16/May/21

z = f (x, y) = {12 x}^{- 1} + {18 y}^{- 1} + xy

\Rightarrow z_{x} = - \frac{12}{x^{2}} + y = 0, y = \frac{12}{x^{2}} \dots (i)

\Rightarrow z_{y} = - \frac{18}{y^{2}} + x = 0, x = \frac{18}{y^{2}} \dots (ii)

\Rightarrow \frac{12}{18} = \frac{x^{2} y}{{xy}^{2}}; \frac{2}{3} = \frac{x}{y} or y = \frac{3 x}{2}

\Rightarrow \frac{3 x}{2} = \frac{12}{x^{2}} \Rightarrow x = \sqrt[3]{8} = 2 & y = 3

minimum z = f (2, 3) = 18 ⋇

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