Question and Answers Forum

All Questions      Topic List

Mensuration Questions

Previous in All Question      Next in All Question      

Previous in Mensuration      Next in Mensuration      

Question Number 81564 by jagoll last updated on 14/Feb/20

given ((a+b)/c) + ((a+c)/b) +((b+c)/a) = 9 a^2 +b^2 +c^2 = 12 maximum value of a+b+c is

$${given}\: \\ $$$$\frac{{a}+{b}}{{c}}\:+\:\frac{{a}+{c}}{{b}}\:+\frac{{b}+{c}}{{a}}\:=\:\mathrm{9} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:=\:\mathrm{12}\: \\ $$$${maximum}\:{value}\:{of}\:{a}+{b}+{c}\:{is} \\ $$

Commented by mr W last updated on 14/Feb/20

a+b+c≤6(√3) ?

$${a}+{b}+{c}\leqslant\mathrm{6}\sqrt{\mathrm{3}}\:? \\ $$

Commented by john santu last updated on 14/Feb/20

let a+b+c = k ((a+b+c−c)/c)+((a+b+c−b)/b)+((a+b+c−a)/a)=9 (k/c)+(k/b)+(k/a)=12⇒ k(((b+c)/(bc))+(1/a))= 12 k(((ab+bc+ac)/(abc)))=12 let ab+bc+ac=m mk = 12abc (i) a^2 +b^2 +c^2 = (a+b+c)^2 −2(ab+ac+bc) ab+ac+bc = ((k^2 −12)/2) (ii)k(((k^2 −12)/(2abc)))=12 ⇒ k^3 −12k=24abc≤24((k/3))^3 27k^3 −324k≤24k^3 3k(k^2 −108)≤0 3k (k+6(√3))(k−6(√3))≤0 for k> 0 ⇒0 < k ≤ 6(√3)

$$\mathrm{let}\:\mathrm{a}+\mathrm{b}+\mathrm{c}\:=\:\mathrm{k} \\ $$$$\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}−\mathrm{c}}{\mathrm{c}}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}−\mathrm{b}}{\mathrm{b}}+\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}−\mathrm{a}}{\mathrm{a}}=\mathrm{9} \\ $$$$\frac{\mathrm{k}}{\mathrm{c}}+\frac{\mathrm{k}}{\mathrm{b}}+\frac{\mathrm{k}}{\mathrm{a}}=\mathrm{12}\Rightarrow\: \\ $$$$\mathrm{k}\left(\frac{\mathrm{b}+\mathrm{c}}{\mathrm{bc}}+\frac{\mathrm{1}}{\mathrm{a}}\right)=\:\mathrm{12} \\ $$$$\mathrm{k}\left(\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ac}}{\mathrm{abc}}\right)=\mathrm{12} \\ $$$$\mathrm{let}\:\mathrm{ab}+\mathrm{bc}+\mathrm{ac}=\mathrm{m} \\ $$$$\mathrm{mk}\:=\:\mathrm{12abc} \\ $$$$\left(\mathrm{i}\right)\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} \:=\:\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\right) \\ $$$$\mathrm{ab}+\mathrm{ac}+\mathrm{bc}\:=\:\frac{\mathrm{k}^{\mathrm{2}} −\mathrm{12}}{\mathrm{2}} \\ $$$$\left(\mathrm{ii}\right)\mathrm{k}\left(\frac{\mathrm{k}^{\mathrm{2}} −\mathrm{12}}{\mathrm{2abc}}\right)=\mathrm{12}\:\Rightarrow\:\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\mathrm{24}\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} \\ $$$$\mathrm{27k}^{\mathrm{3}} −\mathrm{324k}\leqslant\mathrm{24k}^{\mathrm{3}} \\ $$$$\mathrm{3k}\left(\mathrm{k}^{\mathrm{2}} −\mathrm{108}\right)\leqslant\mathrm{0}\: \\ $$$$\mathrm{3k}\:\left(\mathrm{k}+\mathrm{6}\sqrt{\mathrm{3}}\right)\left(\mathrm{k}−\mathrm{6}\sqrt{\mathrm{3}}\right)\leqslant\mathrm{0} \\ $$$$\mathrm{for}\:\mathrm{k}>\:\mathrm{0}\:\Rightarrow\mathrm{0}\:<\:\mathrm{k}\:\leqslant\:\mathrm{6}\sqrt{\mathrm{3}}\: \\ $$

Commented by mr W last updated on 14/Feb/20

you need only a little step to the final solution: k^3 −12k=24abc≤24(((a+b+c)/3))^3 =((24k^3 )/(27)) ⇒(k^3 /9)−12k≤0 ⇒k≤(√(9×12))=6(√3)

$${you}\:{need}\:{only}\:{a}\:{little}\:{step}\:{to}\:{the}\:{final} \\ $$$${solution}: \\ $$$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\mathrm{24}\left(\frac{{a}+{b}+{c}}{\mathrm{3}}\right)^{\mathrm{3}} =\frac{\mathrm{24}{k}^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\Rightarrow\frac{{k}^{\mathrm{3}} }{\mathrm{9}}−\mathrm{12k}\leqslant\mathrm{0} \\ $$$$\Rightarrow{k}\leqslant\sqrt{\mathrm{9}×\mathrm{12}}=\mathrm{6}\sqrt{\mathrm{3}} \\ $$

Commented by mr W last updated on 14/Feb/20

please discuss: can a+b+c reach this maximum 6(√3)? i mean, is a+b+c<6(√3) or a+b+c≤6(√3) correct?

$${please}\:{discuss}: \\ $$$${can}\:{a}+{b}+{c}\:{reach}\:{this}\:{maximum}\:\mathrm{6}\sqrt{\mathrm{3}}? \\ $$$${i}\:{mean},\:{is}\:{a}+{b}+{c}<\mathrm{6}\sqrt{\mathrm{3}}\:{or} \\ $$$${a}+{b}+{c}\leqslant\mathrm{6}\sqrt{\mathrm{3}}\:{correct}? \\ $$

Commented by john santu last updated on 14/Feb/20

we use AM sir? yes i agree with you

$$\mathrm{we}\:\mathrm{use}\:\mathrm{AM}\:\mathrm{sir}?\:\mathrm{yes}\:\mathrm{i}\:\mathrm{agree}\:\mathrm{with}\:\mathrm{you} \\ $$

Commented by mr W last updated on 14/Feb/20

my question is if the “=” sign in a+b+c≤6(√3) is valid. i think it isn′t. i.e. correct is: a+b+c<6(√3). what′s your opinion?

$${my}\:{question}\:{is}\:{if}\:{the}\:``=''\:{sign}\:{in} \\ $$$${a}+{b}+{c}\leqslant\mathrm{6}\sqrt{\mathrm{3}}\:{is}\:{valid}.\:\:{i}\:{think}\:{it}\:{isn}'{t}. \\ $$$${i}.{e}.\:{correct}\:{is}: \\ $$$${a}+{b}+{c}<\mathrm{6}\sqrt{\mathrm{3}}. \\ $$$${what}'{s}\:{your}\:{opinion}? \\ $$

Commented by jagoll last updated on 14/Feb/20

i not find this solution sir.

$${i}\:{not}\:{find}\:{this}\:{solution}\:{sir}. \\ $$

Commented by mr W last updated on 14/Feb/20

you mean this solution is wrong or you mean you don′t know the correct solution?

$${you}\:{mean}\:{this}\:{solution}\:{is}\:{wrong}\:{or} \\ $$$${you}\:{mean}\:{you}\:{don}'{t}\:{know}\:{the}\:{correct} \\ $$$${solution}? \\ $$

Commented by jagoll last updated on 14/Feb/20

i think this solution is correct. but i don′t know the solution does it match in the book? answer in the book were not included

$${i}\:{think}\:{this}\:{solution}\:{is}\:{correct}. \\ $$$${but}\:{i}\:{don}'{t}\:{know}\:{the}\:{solution}\:{does} \\ $$$${it}\:{match}\:{in}\:{the}\:{book}?\:{answer} \\ $$$${in}\:{the}\:{book}\:{were}\:{not}\:{included} \\ $$

Commented by mr W last updated on 14/Feb/20

k^3 −12k=24abc≤ ((k/3))^3 is wrong. correct is: k^3 −12k=24abc≤ 24((k/3))^3 . this is for k>0. if k<0, then k^3 −12k=24abc≥−24((k/3))^3 .

$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\:\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} \:{is}\:{wrong}. \\ $$$${correct}\:{is}: \\ $$$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\leqslant\:\mathrm{24}\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} . \\ $$$${this}\:{is}\:{for}\:{k}>\mathrm{0}. \\ $$$$ \\ $$$${if}\:{k}<\mathrm{0},\:{then} \\ $$$$\mathrm{k}^{\mathrm{3}} −\mathrm{12k}=\mathrm{24abc}\geqslant−\mathrm{24}\left(\frac{\mathrm{k}}{\mathrm{3}}\right)^{\mathrm{3}} . \\ $$

Commented by john santu last updated on 14/Feb/20

oo yes it my typo

$$\mathrm{oo}\:\mathrm{yes}\:\mathrm{it}\:\mathrm{my}\:\mathrm{typo} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com