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Question-151357

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Question Number 151357 by mathdanisur last updated on 20/Aug/21
Commented by mr W last updated on 20/Aug/21
check the question please! Σ_(k=1) ^(10) a_k ≥100 and Σ_(k=1) ^(10) a_k ^2 ≤1001 ???
$${check}\:{the}\:{question}\:{please}! \\ $$$$\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{a}_{{k}} \geqslant\mathrm{100}\:{and}\:\underset{{k}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}{a}_{{k}} ^{\mathrm{2}} \leqslant\mathrm{1001}\:\:\:??? \\ $$
Commented by mathdanisur last updated on 20/Aug/21
Sorry Ser yes Σ_( k=1) ^(10) a_k ^2 ≤ 1001
$$\mathrm{Sorry}\:\mathrm{Ser}\:\mathrm{yes}\:\underset{\:\boldsymbol{\mathrm{k}}=\mathrm{1}} {\overset{\mathrm{10}} {\sum}}\boldsymbol{\mathrm{a}}_{\boldsymbol{\mathrm{k}}} ^{\mathrm{2}} \:\leqslant\:\mathrm{1001} \\ $$
Commented by mr W last updated on 20/Aug/21
min a_k =10−(3/( (√(10)))) max a_k =10+(3/( (√(10))))
$${min}\:{a}_{{k}} =\mathrm{10}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${max}\:{a}_{{k}} =\mathrm{10}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$
Commented by mathdanisur last updated on 20/Aug/21
Thank You Dear Ser if possible pkease
$$\mathrm{Thank}\:\mathrm{You}\:\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$$$\mathrm{if}\:\mathrm{possible}\:\mathrm{pkease} \\ $$
Commented by mr W last updated on 20/Aug/21
what′s the answer in your book?
$${what}'{s}\:{the}\:{answer}\:{in}\:{your}\:{book}? \\ $$
Commented by mathdanisur last updated on 20/Aug/21
Yes Dear Ser, the answer is corrcet solution please Ser
$$\mathrm{Yes}\:\mathrm{Dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{corrcet} \\ $$$$\mathrm{solution}\:\mathrm{please}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$
Answered by mr W last updated on 20/Aug/21
the mean of the 10 numbers is 10. say the smallest number is 10−a, such that this number is as small as possible, the other nine numbers should be bigger than 10 and as big as possible. such that Σa_k ^2 is as small as possible, these nine numbers should be as small as possible. this can only be reached if the nine numbers are equal. so we have one number 10−a, the other nine ones are larger than or equal to 10+(a/9). S=Σa_k ^2 ≥(10−a)^2 +9×(10+(a/9))^2 since S≤1001, 1001≥(10−a)^2 +9×(10+(a/9))^2 ⇒9≥10a^2 ⇒a≤(3/( (√(10)))) the smallest number is 10−a≥10−(3/( (√(10)))) that means the smallest number is 10−(3/( (√(10)))). similarly, say the largest number is 10+b, then the other nine numbers should be equal and equal to 10−(b/9). 1001≥S=Σa_k ^2 ≥(10+b)^2 +9×(10−(a/9))^2 ⇒9≥10b^2 ⇒b≤(3/( (√(10)))) the largest number is 10+b≤10+(3/( (√(10)))) that means the largest number is 10+(3/( (√(10)))).
$${the}\:{mean}\:{of}\:{the}\:\mathrm{10}\:{numbers}\:{is}\:\mathrm{10}. \\ $$$${say}\:{the}\:{smallest}\:{number}\:{is}\:\mathrm{10}−{a}, \\ $$$${such}\:{that}\:{this}\:{number}\:{is}\:{as}\:{small} \\ $$$${as}\:{possible},\:{the}\:{other}\:{nine}\:{numbers} \\ $$$${should}\:{be}\:{bigger}\:{than}\:\mathrm{10}\:{and}\:{as}\:{big} \\ $$$${as}\:{possible}.\:{such}\:{that}\:\Sigma{a}_{{k}} ^{\mathrm{2}} \:{is}\:{as}\:{small} \\ $$$${as}\:{possible},\:{these}\:{nine}\:{numbers}\: \\ $$$${should}\:{be}\:{as}\:{small}\:{as}\:{possible}.\:{this} \\ $$$${can}\:{only}\:{be}\:{reached}\:{if}\:{the}\:{nine} \\ $$$${numbers}\:{are}\:{equal}.\:{so}\:{we}\:{have}\:{one} \\ $$$${number}\:\mathrm{10}−{a},\:{the}\:{other}\:{nine}\:{ones} \\ $$$${are}\:{larger}\:{than}\:{or}\:{equal}\:{to}\:\mathrm{10}+\frac{{a}}{\mathrm{9}}. \\ $$$${S}=\Sigma{a}_{{k}} ^{\mathrm{2}} \geqslant\left(\mathrm{10}−{a}\right)^{\mathrm{2}} +\mathrm{9}×\left(\mathrm{10}+\frac{{a}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$${since}\:{S}\leqslant\mathrm{1001}, \\ $$$$\mathrm{1001}\geqslant\left(\mathrm{10}−{a}\right)^{\mathrm{2}} +\mathrm{9}×\left(\mathrm{10}+\frac{{a}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}\geqslant\mathrm{10}{a}^{\mathrm{2}} \\ $$$$\Rightarrow{a}\leqslant\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${the}\:{smallest}\:{number}\:{is} \\ $$$$\mathrm{10}−{a}\geqslant\mathrm{10}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${that}\:{means}\:{the}\:{smallest}\:{number} \\ $$$${is}\:\mathrm{10}−\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}. \\ $$$${similarly},\:{say}\:{the}\:{largest}\:{number} \\ $$$${is}\:\mathrm{10}+{b},\:{then}\:{the}\:{other}\:{nine}\:{numbers} \\ $$$${should}\:{be}\:{equal}\:{and}\:{equal}\:{to}\:\mathrm{10}−\frac{{b}}{\mathrm{9}}. \\ $$$$\mathrm{1001}\geqslant{S}=\Sigma{a}_{{k}} ^{\mathrm{2}} \geqslant\left(\mathrm{10}+{b}\right)^{\mathrm{2}} +\mathrm{9}×\left(\mathrm{10}−\frac{{a}}{\mathrm{9}}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{9}\geqslant\mathrm{10}{b}^{\mathrm{2}} \\ $$$$\Rightarrow{b}\leqslant\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${the}\:{largest}\:{number}\:{is} \\ $$$$\mathrm{10}+{b}\leqslant\mathrm{10}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}} \\ $$$${that}\:{means}\:{the}\:{largest}\:{number} \\ $$$${is}\:\mathrm{10}+\frac{\mathrm{3}}{\:\sqrt{\mathrm{10}}}. \\ $$
Commented by mathdanisur last updated on 20/Aug/21
Nice Ser, Thank You
$$\mathrm{Nice}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{Thank}\:\mathrm{You} \\ $$

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