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Question Number 134817 by bramlexs22 last updated on 07/Mar/21
What is the maximum value of abc if a+b+c = 5 and ab+bc+ca = 7?
$$ \\ $$What is the maximum value of abc if a+b+c = 5 and ab+bc+ca = 7?
Commented by bramlexs22 last updated on 09/Mar/21
thanks you all sir
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{all}\:\mathrm{sir} \\ $$
Commented by john_santu last updated on 08/Mar/21
let a+b = x then c = 5−x (i) xc = (a+b)c = ac+bc = 7−ab then ab = 7−xc so abc = (7−xc)c abc = 7c−xc^2 , injecting c=5−x we get abc = f(x)= 7(5−x)−x(5−x)^2 f(x)= 35−7x−x(x^2 −10x+25) f(x)=−x^3 +10x^2 −32x+35 (df/dx) = −3x^2 +20x−32 = 0 we get { ((x=(8/3))),((x=4)) :} (d^2 f/dx^2 ) = −6x+20∣_(x= 4) <0 and (d^2 f/dx^2 ) = −6x+20∣_(x = (8/3)) = −((48)/3)+20 >0 so f(x)_(max) for x = 4 and f(x)_(min ) for x=(8/3) ∴ max value is f(4)=−4^3 +10.4^2 −32.4+35 f(4)= −64+160−128+35 = 3 when a+b=4 and c = 1 . from eq(2) ab+ac+bc = 7⇒ab+a+b = 7 ; ab = 3 ∧b = 4−a ; we get a(4−a)=3 a^2 −4a+3 = 0 → { ((a=1→b=3 or)),((a=3→b=1 )) :} (a,b,c) = (1,3,1) or (3,1,1) clearly max is f(4) = 3 min value is f((8/3))=−((8/3))^3 +10((8/3))^2 −32((8/3))+35
$${let}\:{a}+{b}\:=\:{x}\:{then}\:{c}\:=\:\mathrm{5}−{x} \\ $$$$\left({i}\right)\:{xc}\:=\:\left({a}+{b}\right){c}\:=\:{ac}+{bc}\:=\:\mathrm{7}−{ab} \\ $$$${then}\:{ab}\:=\:\mathrm{7}−{xc}\:{so}\:{abc}\:=\:\left(\mathrm{7}−{xc}\right){c} \\ $$$${abc}\:=\:\mathrm{7}{c}−{xc}^{\mathrm{2}} \:,\:{injecting}\:{c}=\mathrm{5}−{x} \\ $$$${we}\:{get}\:{abc}\:=\:{f}\left({x}\right)=\:\mathrm{7}\left(\mathrm{5}−{x}\right)−{x}\left(\mathrm{5}−{x}\right)^{\mathrm{2}} \\ $$$${f}\left({x}\right)=\:\mathrm{35}−\mathrm{7}{x}−{x}\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{25}\right) \\ $$$${f}\left({x}\right)=−{x}^{\mathrm{3}} +\mathrm{10}{x}^{\mathrm{2}} −\mathrm{32}{x}+\mathrm{35}\: \\ $$$$\frac{{df}}{{dx}}\:=\:−\mathrm{3}{x}^{\mathrm{2}} +\mathrm{20}{x}−\mathrm{32}\:=\:\mathrm{0} \\ $$$${we}\:{get}\:\begin{cases}{{x}=\frac{\mathrm{8}}{\mathrm{3}}}\\{{x}=\mathrm{4}}\end{cases} \\ $$$$\:\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }\:=\:−\mathrm{6}{x}+\mathrm{20}\mid_{{x}=\:\mathrm{4}} <\mathrm{0}\:\:{and}\: \\ $$$$\frac{{d}^{\mathrm{2}} {f}}{{dx}^{\mathrm{2}} }\:=\:−\mathrm{6}{x}+\mathrm{20}\mid_{{x}\:=\:\frac{\mathrm{8}}{\mathrm{3}}} =\:−\frac{\mathrm{48}}{\mathrm{3}}+\mathrm{20}\:>\mathrm{0}\: \\ $$$${so}\:{f}\left({x}\right)_{{max}} \:{for}\:{x}\:=\:\mathrm{4}\:{and}\:{f}\left({x}\right)_{{min}\:} {for}\:{x}=\frac{\mathrm{8}}{\mathrm{3}} \\ $$$$\therefore\:{max}\:{value}\:{is}\:{f}\left(\mathrm{4}\right)=−\mathrm{4}^{\mathrm{3}} +\mathrm{10}.\mathrm{4}^{\mathrm{2}} −\mathrm{32}.\mathrm{4}+\mathrm{35}\: \\ $$$${f}\left(\mathrm{4}\right)=\:−\mathrm{64}+\mathrm{160}−\mathrm{128}+\mathrm{35}\:=\:\mathrm{3} \\ $$$${when}\:{a}+{b}=\mathrm{4}\:{and}\:{c}\:=\:\mathrm{1}\:.\:{from}\:{eq}\left(\mathrm{2}\right) \\ $$$${ab}+{ac}+{bc}\:=\:\mathrm{7}\Rightarrow{ab}+{a}+{b}\:=\:\mathrm{7}\:;\: \\ $$$${ab}\:=\:\mathrm{3}\:\wedge{b}\:=\:\mathrm{4}−{a}\:;\:{we}\:{get}\:{a}\left(\mathrm{4}−{a}\right)=\mathrm{3} \\ $$$${a}^{\mathrm{2}} −\mathrm{4}{a}+\mathrm{3}\:=\:\mathrm{0}\:\rightarrow\begin{cases}{{a}=\mathrm{1}\rightarrow{b}=\mathrm{3}\:{or}}\\{{a}=\mathrm{3}\rightarrow{b}=\mathrm{1}\:}\end{cases} \\ $$$$\left({a},{b},{c}\right)\:=\:\left(\mathrm{1},\mathrm{3},\mathrm{1}\right)\:{or}\:\left(\mathrm{3},\mathrm{1},\mathrm{1}\right) \\ $$$${clearly}\:{max}\:{is}\:{f}\left(\mathrm{4}\right)\:=\:\mathrm{3} \\ $$$${min}\:{value}\:{is}\:{f}\left(\frac{\mathrm{8}}{\mathrm{3}}\right)=−\left(\frac{\mathrm{8}}{\mathrm{3}}\right)^{\mathrm{3}} +\mathrm{10}\left(\frac{\mathrm{8}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{32}\left(\frac{\mathrm{8}}{\mathrm{3}}\right)+\mathrm{35} \\ $$$$ \\ $$
Commented by mr W last updated on 08/Mar/21
very nice! i used also similar method, but maybe made a calculation mistake somewhere.
$${very}\:{nice}!\:{i}\:{used}\:{also}\:{similar}\:{method},\: \\ $$$${but}\:{maybe}\:{made}\:{a}\:{calculation}\:{mistake} \\ $$$${somewhere}. \\ $$
Answered by liberty last updated on 07/Mar/21
(1)a=5−(b+c) (2) (5−(b+c))b+bc+c(5−(b+c))=7 ⇒ 5b−b^2 −bc+bc+5c−bc−c^2 =7 ⇒5b−b^2 −bc+5c+c^2 = 7 (3)abc = (5−(b+c))bc =5bc−b^2 c−bc^2 let f(b,c)=5bc−b^2 c−bc^2 +λ(5b−b^2 −bc+5c+c^2 −7) (i) f_b = 5c−2bc−c^2 +λ(5−2b−c)=0 (ii)f_c = 5b−b^2 −2bc+λ(−b+5+2c)= 0 (iii)f_(λ ) = 5b−b^2 −bc+5c+c^2 = 7 from eq (i) &(ii) we get { ((λ=((c^2 +2bc−5c)/(5−2b−c)))),((λ = ((b^2 +2bc−5b)/(5−b+2c)))) :} and b^2 −5b = c^2 +5c−bc−7 ⇒ λ = λ ; ((c^2 +2bc−5c)/(5−2b−c)) = ((c^2 +5c−bc−7+2bc)/(5−b+2c)) next...
$$\left(\mathrm{1}\right){a}=\mathrm{5}−\left({b}+{c}\right)\:\left(\mathrm{2}\right)\:\left(\mathrm{5}−\left({b}+{c}\right)\right){b}+{bc}+{c}\left(\mathrm{5}−\left({b}+{c}\right)\right)=\mathrm{7} \\ $$$$\Rightarrow\:\mathrm{5}{b}−{b}^{\mathrm{2}} −{bc}+{bc}+\mathrm{5}{c}−{bc}−{c}^{\mathrm{2}} =\mathrm{7} \\ $$$$\Rightarrow\mathrm{5}{b}−{b}^{\mathrm{2}} −{bc}+\mathrm{5}{c}+{c}^{\mathrm{2}} \:=\:\mathrm{7} \\ $$$$\left(\mathrm{3}\right){abc}\:=\:\left(\mathrm{5}−\left({b}+{c}\right)\right){bc}\:=\mathrm{5}{bc}−{b}^{\mathrm{2}} {c}−{bc}^{\mathrm{2}} \\ $$$${let}\:{f}\left({b},{c}\right)=\mathrm{5}{bc}−{b}^{\mathrm{2}} {c}−{bc}^{\mathrm{2}} +\lambda\left(\mathrm{5}{b}−{b}^{\mathrm{2}} −{bc}+\mathrm{5}{c}+{c}^{\mathrm{2}} −\mathrm{7}\right) \\ $$$$\left({i}\right)\:{f}_{{b}} =\:\mathrm{5}{c}−\mathrm{2}{bc}−{c}^{\mathrm{2}} +\lambda\left(\mathrm{5}−\mathrm{2}{b}−{c}\right)=\mathrm{0} \\ $$$$\left({ii}\right){f}_{{c}} =\:\mathrm{5}{b}−{b}^{\mathrm{2}} −\mathrm{2}{bc}+\lambda\left(−{b}+\mathrm{5}+\mathrm{2}{c}\right)=\:\mathrm{0} \\ $$$$\left({iii}\right){f}_{\lambda\:} =\:\mathrm{5}{b}−{b}^{\mathrm{2}} −{bc}+\mathrm{5}{c}+{c}^{\mathrm{2}} \:=\:\mathrm{7} \\ $$$${from}\:{eq}\:\left({i}\right)\:\&\left({ii}\right)\:{we}\:{get}\:\begin{cases}{\lambda=\frac{{c}^{\mathrm{2}} +\mathrm{2}{bc}−\mathrm{5}{c}}{\mathrm{5}−\mathrm{2}{b}−{c}}}\\{\lambda\:=\:\frac{{b}^{\mathrm{2}} +\mathrm{2}{bc}−\mathrm{5}{b}}{\mathrm{5}−{b}+\mathrm{2}{c}}}\end{cases} \\ $$$${and}\:{b}^{\mathrm{2}} −\mathrm{5}{b}\:=\:{c}^{\mathrm{2}} +\mathrm{5}{c}−{bc}−\mathrm{7} \\ $$$$\Rightarrow\:\lambda\:=\:\lambda\:;\:\frac{{c}^{\mathrm{2}} +\mathrm{2}{bc}−\mathrm{5}{c}}{\mathrm{5}−\mathrm{2}{b}−{c}}\:=\:\frac{{c}^{\mathrm{2}} +\mathrm{5}{c}−{bc}−\mathrm{7}+\mathrm{2}{bc}}{\mathrm{5}−{b}+\mathrm{2}{c}} \\ $$$${next}… \\ $$
Answered by mr W last updated on 08/Mar/21
c=5−(a+b) ab+(a+b)c=7 c=((7−ab)/(a+b))=5−(a+b) ⇒ab=7−(a+b)(5−(a+b)) P=abc=ab(5−(a+b))=[7−(a+b)(5−(a+b))](5−(a+b)) let t=a+b ⇒P=[7−t(5−t)](5−t)=−t^3 +10t^2 −32t+35 (dP/dt)=−3t^2 +20t−32=0 (3t−8)(t−4)=0 ⇒t=(8/3), 4 (d^2 P/dt^2 )=−6t+20 at t=(8/3): (d^2 P/dt^2 )=−6×(8/3)+20=4>0 ⇒min. at t=4: (d^2 P/dt^2 )=−6×4+20=−4<0 ⇒max. ⇒P_(min) =[7−(8/3)(5−(8/3))](5−(8/3))=((49)/(27)) ⇒P_(max) =[7−4(5−4)](5−4)=3
$${c}=\mathrm{5}−\left({a}+{b}\right) \\ $$$${ab}+\left({a}+{b}\right){c}=\mathrm{7} \\ $$$${c}=\frac{\mathrm{7}−{ab}}{{a}+{b}}=\mathrm{5}−\left({a}+{b}\right) \\ $$$$\Rightarrow{ab}=\mathrm{7}−\left({a}+{b}\right)\left(\mathrm{5}−\left({a}+{b}\right)\right) \\ $$$${P}={abc}={ab}\left(\mathrm{5}−\left({a}+{b}\right)\right)=\left[\mathrm{7}−\left({a}+{b}\right)\left(\mathrm{5}−\left({a}+{b}\right)\right)\right]\left(\mathrm{5}−\left({a}+{b}\right)\right) \\ $$$${let}\:{t}={a}+{b} \\ $$$$\Rightarrow{P}=\left[\mathrm{7}−{t}\left(\mathrm{5}−{t}\right)\right]\left(\mathrm{5}−{t}\right)=−{t}^{\mathrm{3}} +\mathrm{10}{t}^{\mathrm{2}} −\mathrm{32}{t}+\mathrm{35} \\ $$$$\frac{{dP}}{{dt}}=−\mathrm{3}{t}^{\mathrm{2}} +\mathrm{20}{t}−\mathrm{32}=\mathrm{0} \\ $$$$\left(\mathrm{3}{t}−\mathrm{8}\right)\left({t}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\Rightarrow{t}=\frac{\mathrm{8}}{\mathrm{3}},\:\mathrm{4} \\ $$$$\frac{{d}^{\mathrm{2}} {P}}{{dt}^{\mathrm{2}} }=−\mathrm{6}{t}+\mathrm{20} \\ $$$${at}\:{t}=\frac{\mathrm{8}}{\mathrm{3}}:\:\frac{{d}^{\mathrm{2}} {P}}{{dt}^{\mathrm{2}} }=−\mathrm{6}×\frac{\mathrm{8}}{\mathrm{3}}+\mathrm{20}=\mathrm{4}>\mathrm{0}\:\Rightarrow{min}. \\ $$$${at}\:{t}=\mathrm{4}:\:\frac{{d}^{\mathrm{2}} {P}}{{dt}^{\mathrm{2}} }=−\mathrm{6}×\mathrm{4}+\mathrm{20}=−\mathrm{4}<\mathrm{0}\:\Rightarrow{max}. \\ $$$$\Rightarrow{P}_{{min}} =\left[\mathrm{7}−\frac{\mathrm{8}}{\mathrm{3}}\left(\mathrm{5}−\frac{\mathrm{8}}{\mathrm{3}}\right)\right]\left(\mathrm{5}−\frac{\mathrm{8}}{\mathrm{3}}\right)=\frac{\mathrm{49}}{\mathrm{27}} \\ $$$$\Rightarrow{P}_{{max}} =\left[\mathrm{7}−\mathrm{4}\left(\mathrm{5}−\mathrm{4}\right)\right]\left(\mathrm{5}−\mathrm{4}\right)=\mathrm{3} \\ $$
Answered by ajfour last updated on 08/Mar/21
x^3 −5x^2 +7x−m=0 where a+b+c=5 ab+bc+ca=7 abc=m m=x^3 −5x^2 +7x (dm/dx)=3x^2 −10x+7 x=((10±(√(100−84)))/6) x=((10±4)/6) ⇒ x=1, (7/3) for x=1, m=3 for x=(7/3) ,m=((49)/(27)) hence (abc)_(max) = m_(max) =3
$${x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{7}{x}−{m}=\mathrm{0} \\ $$$${where} \\ $$$${a}+{b}+{c}=\mathrm{5} \\ $$$${ab}+{bc}+{ca}=\mathrm{7} \\ $$$${abc}={m} \\ $$$${m}={x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{7}{x} \\ $$$$\frac{{dm}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{7} \\ $$$${x}=\frac{\mathrm{10}\pm\sqrt{\mathrm{100}−\mathrm{84}}}{\mathrm{6}} \\ $$$${x}=\frac{\mathrm{10}\pm\mathrm{4}}{\mathrm{6}}\:\:\Rightarrow\:\:{x}=\mathrm{1},\:\frac{\mathrm{7}}{\mathrm{3}} \\ $$$${for}\:{x}=\mathrm{1},\:\:{m}=\mathrm{3} \\ $$$${for}\:\:{x}=\frac{\mathrm{7}}{\mathrm{3}}\:\:,{m}=\frac{\mathrm{49}}{\mathrm{27}} \\ $$$${hence}\:\left({abc}\right)_{{max}} =\:{m}_{{max}} =\mathrm{3} \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 09/Mar/21
nice idea!
$${nice}\:{idea}! \\ $$
Commented by ajfour last updated on 09/Mar/21
thanks sir, m little shaken by circumstances, n anyway taken to watching Utube movies n music, cant focus here like i used to, lets see how long; I m so very unpredictable even to myself...
$${thanks}\:{sir},\:{m}\:{little}\:{shaken}\:{by} \\ $$$${circumstances},\:{n}\:{anyway}\:{taken} \\ $$$${to}\:{watching}\:{Utube}\:{movies}\:{n} \\ $$$${music},\:{cant}\:{focus}\:{here}\:{like}\:{i} \\ $$$${used}\:{to},\:{lets}\:{see}\:{how}\:{long};\:{I}\:{m} \\ $$$${so}\:{very}\:{unpredictable}\:{even}\:{to} \\ $$$${myself}… \\ $$
Commented by mr W last updated on 09/Mar/21
take best care of yourself sir!
$${take}\:{best}\:{care}\:{of}\:{yourself}\:{sir}! \\ $$

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