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Question-211946

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Question Number 211946 by BaliramKumar last updated on 25/Sep/24
Answered by BHOOPENDRA last updated on 25/Sep/24
(2+3+4)^2 =9^2 =81
$$\left(\mathrm{2}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} =\mathrm{9}^{\mathrm{2}} =\mathrm{81} \\ $$
Commented by BHOOPENDRA last updated on 25/Sep/24
method(II) from Couchy schwarz inequality ((4/x)+(9/y)+((16)/z))(x+y+z)≥ (2+3+4)^2 =81 ((4/x)+(9/y)+((16)/z))≥81
$${method}\left({II}\right) \\ $$$${from}\:{Couchy}\:{schwarz}\:{inequality} \\ $$$$\:\left(\frac{\mathrm{4}}{{x}}+\frac{\mathrm{9}}{{y}}+\frac{\mathrm{16}}{{z}}\right)\left({x}+{y}+{z}\right)\geq\:\left(\mathrm{2}+\mathrm{3}+\mathrm{4}\right)^{\mathrm{2}} \\ $$$$=\mathrm{81} \\ $$$$\left(\frac{\mathrm{4}}{{x}}+\frac{\mathrm{9}}{{y}}+\frac{\mathrm{16}}{{z}}\right)\geq\mathrm{81} \\ $$
Answered by BHOOPENDRA last updated on 25/Sep/24
there are many method to find minimum value of that f(x,y,z)=x+y+z =1 (1) g(x,y,z)=(4/x)+(9/y)+((16)/z) ▽f(x,y,z)= λ▽g(x,y,z) ▽f=(1,1,1) ▽g=(−(4/x^2 ) ,((−9)/y^2 ) ,((−16)/z^2 )) (1,1,1)=(((−4λ)/x^2 ), ((−9λ)/y^2 ),((−16λ)/z^2 )) compare 1=((−4λ)/x^2 ), x^2 =−4λ ,let (−λ)=p x=2(√(p )) ,similarly y=3(√p) ,z=4(√p) put the value x+y+z=1 2(√p) +3(√p) +4(√p) =1 (√p) =(1/9) p=(1/(81)) (4/(2×1/9))+(9/(3×1/9))+((16)/(4×1/9)) =18+27+36=81
$${there}\:{are}\:{many}\:{method}\:{to}\:{find}\:{minimum} \\ $$$${value}\:{of}\:{that} \\ $$$${f}\left({x},{y},{z}\right)={x}+{y}+{z}\:=\mathrm{1}\:\:\left(\mathrm{1}\right) \\ $$$${g}\left({x},{y},{z}\right)=\frac{\mathrm{4}}{{x}}+\frac{\mathrm{9}}{{y}}+\frac{\mathrm{16}}{{z}} \\ $$$$\bigtriangledown{f}\left({x},{y},{z}\right)=\:\lambda\bigtriangledown{g}\left({x},{y},{z}\right) \\ $$$$\bigtriangledown{f}=\left(\mathrm{1},\mathrm{1},\mathrm{1}\right) \\ $$$$\bigtriangledown{g}=\left(−\frac{\mathrm{4}}{{x}^{\mathrm{2}} }\:,\frac{−\mathrm{9}}{{y}^{\mathrm{2}} }\:,\frac{−\mathrm{16}}{{z}^{\mathrm{2}} }\right) \\ $$$$\left(\mathrm{1},\mathrm{1},\mathrm{1}\right)=\left(\frac{−\mathrm{4}\lambda}{{x}^{\mathrm{2}} },\:\frac{−\mathrm{9}\lambda}{{y}^{\mathrm{2}} },\frac{−\mathrm{16}\lambda}{{z}^{\mathrm{2}} }\right) \\ $$$${compare} \\ $$$$\mathrm{1}=\frac{−\mathrm{4}\lambda}{{x}^{\mathrm{2}} },\:{x}^{\mathrm{2}} =−\mathrm{4}\lambda\:,{let}\:\left(−\lambda\right)={p} \\ $$$${x}=\mathrm{2}\sqrt{{p}\:}\:,{similarly} \\ $$$${y}=\mathrm{3}\sqrt{{p}}\:,{z}=\mathrm{4}\sqrt{{p}}\: \\ $$$${put}\:{the}\:{value} \\ $$$${x}+{y}+{z}=\mathrm{1} \\ $$$$\mathrm{2}\sqrt{{p}}\:+\mathrm{3}\sqrt{{p}}\:+\mathrm{4}\sqrt{{p}}\:=\mathrm{1} \\ $$$$\sqrt{{p}}\:=\frac{\mathrm{1}}{\mathrm{9}} \\ $$$${p}=\frac{\mathrm{1}}{\mathrm{81}} \\ $$$$ \\ $$$$\frac{\mathrm{4}}{\mathrm{2}×\mathrm{1}/\mathrm{9}}+\frac{\mathrm{9}}{\mathrm{3}×\mathrm{1}/\mathrm{9}}+\frac{\mathrm{16}}{\mathrm{4}×\mathrm{1}/\mathrm{9}} \\ $$$$=\mathrm{18}+\mathrm{27}+\mathrm{36}=\mathrm{81} \\ $$

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