# if-a-b-c-d-e-f-10-and-a-2-b-2-c-2-d-2-e-2-f-2-25-find-a-min-and-f-max-

Question Number 205914 by mr W last updated on 02/Apr/24
$${if}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}\:{and} \\$$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{find} \\$$$${a}_{{min}} \:{and}\:{f}_{{max}} . \\$$
Commented by Tinku Tara last updated on 03/Apr/24
$$\mathrm{Question}:\: \\$$$$\mathrm{Why}\:\mathrm{is}\:\mathrm{min}=−\mathrm{5}\:\mathrm{and}\:\mathrm{max}=+\mathrm{5} \\$$$$\mathrm{not}\:\mathrm{the}\:\mathrm{solution}? \\$$
Commented by mr W last updated on 03/Apr/24
$${if}\:{min}=−\mathrm{5},\:{say}\:{a}=−\mathrm{5},\:{to}\:{fulfill} \\$$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{we}\:{must} \\$$$${have}\:{b}={c}={d}={e}={f}=\mathrm{0},\:{i}.{e}. \\$$$${a}+{b}+{c}+{d}+{e}+{f}=−\mathrm{5},\:{but}\:{this}\:{is} \\$$$${contradiction}\:{to}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}, \\$$$${therefore}\:{a}\neq−\mathrm{5}. \\$$
Answered by mr W last updated on 05/Apr/24
$$\underline{{Method}\:{II}} \\$$$${a}=\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right) \\$$$$\left(\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} −\mathrm{25}=\mathrm{0} \\$$$${F}={a}=\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)+\lambda\left[\left(\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)\right)^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} −\mathrm{25}\right] \\$$$$\frac{\partial{F}}{\partial{b}}=−\mathrm{1}+\lambda\left[−\mathrm{2}\left(\mathrm{10}−\left({b}+{c}+{d}+{e}+{f}\right)\right)+\mathrm{2}{b}\right]=\mathrm{0} \\$$$$=>−\frac{\mathrm{1}}{\mathrm{2}}+\lambda\left[−\mathrm{10}+\left({b}+{c}+{d}+{e}+{f}\right)+{b}\right]=\mathrm{0} \\$$$$\Rightarrow{b}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{10}\lambda−\lambda\left({b}+{c}+{d}+{e}+{f}\right) \\$$$${similarly} \\$$$$\Rightarrow{c}={d}={e}={f}=\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{10}\lambda−\lambda\left({b}+{c}+{d}+{e}+{f}\right) \\$$$${i}.{e}.\:{for}\:{a}_{{min}} \:{or}\:{a}_{{max}} ,\: \\$$$${b}={c}={d}={e}={f}={t},\:{say} \\$$$${then}\:{we}\:{get}\:{as}\:{shown}\:{above} \\$$$${a}_{{min}} =\frac{\mathrm{10}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}},\:{a}_{{max}} =\frac{\mathrm{10}+\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\$$
Answered by mr W last updated on 05/Apr/24
$$\underline{{Method}\:{III}} \\$$$$\overset{\rightarrow} {\boldsymbol{{p}}}=\left({a},{b},{c},{d},{e},{f}\right) \\$$$$\overset{\rightarrow} {\boldsymbol{{q}}}=\left(\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1},\mathrm{1}\right) \\$$$$\mathrm{cos}\:\theta=\frac{\overset{\rightarrow} {\boldsymbol{{p}}}\centerdot\overset{\rightarrow} {\boldsymbol{{q}}}}{\mid\boldsymbol{{p}}\mid\mid\boldsymbol{{q}}\mid} \\$$$$\:\:\:\:=\frac{{a}+{b}+{c}+{d}+{e}+{f}}{\left.\:\sqrt{\mathrm{6}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right.}\right)}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}} \\$$$${a}_{{min}/{max}} =\mathrm{5}\:\mathrm{cos}\:\left(\alpha\pm\theta\right) \\$$$${with}\:\alpha=\mathrm{cos}^{−\mathrm{1}} \frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}} \\$$$${a}_{{min}/{max}} =\mathrm{5}×\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}×\frac{\mathrm{2}}{\:\sqrt{\mathrm{6}}}\mp\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{6}}}×\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{6}}}\right) \\$$$$\:\:\:=\frac{\mathrm{5}\left(\mathrm{2}\mp\sqrt{\mathrm{10}}\right)}{\mathrm{6}}\:\checkmark \\$$
Commented by mr W last updated on 05/Apr/24
Answered by mr W last updated on 04/Apr/24
$$\underline{{Method}\:{I}} \\$$$${if}\:{b}+{c}+{d}+{e}+{f}={s}, \\$$$${b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \geqslant\frac{\left({b}+{c}+{d}+{e}+{f}\right)^{\mathrm{2}} }{\mathrm{5}}=\frac{{s}^{\mathrm{2}} }{\mathrm{5}} \\$$$${equality}\:{holds}\:{when}\:{b}={c}={d}={e}={f}=\frac{{s}}{\mathrm{5}} \\$$$${since}\:{a}+{b}+{c}+{d}+{e}+{f}=\mathrm{10}\:{and} \\$$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} =\mathrm{25},\:{such}\:{that} \\$$$${a}\:{is}\:{as}\:{large}\:{as}\:{possible},\:{b}+{c}+{d}+{e}+{f} \\$$$${should}\:{be}\:{as}\:{small}\:{as}\:{possible}\:{and} \\$$$${besides}\:{b},\:{c},\:{d},\:{e},\:{f}\:{should}\:{be}\:{equal}. \\$$$${say}\:{b}={c}={d}={e}={f}={t},\:{then} \\$$$${a}+\mathrm{5}{t}=\mathrm{10}\:\Rightarrow{t}=\frac{\mathrm{10}−{a}}{\mathrm{5}} \\$$$${a}^{\mathrm{2}} +\mathrm{5}{t}^{\mathrm{2}} =\mathrm{25}\:\Rightarrow{a}^{\mathrm{2}} +\mathrm{5}×\left(\frac{\mathrm{10}−{a}}{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{25} \\$$$$\mathrm{6}{a}^{\mathrm{2}} −\mathrm{20}{a}−\mathrm{25}=\mathrm{0} \\$$$${a}=\frac{\mathrm{10}\pm\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\$$$${i}.{e}.\:{a}_{{min}} =\frac{\mathrm{10}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}},\:{a}_{{max}} =\frac{\mathrm{10}+\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\$$
Answered by mr W last updated on 04/Apr/24
$$\underline{{Method}\:{IV}} \\$$$${f}\left({x}\right)=\left({x}−{b}\right)^{\mathrm{2}} +\left({x}−{c}\right)^{\mathrm{2}} +\left({x}−{d}\right)^{\mathrm{2}} +\left({x}−{e}\right)^{\mathrm{2}} +\left({x}−{f}\right)^{\mathrm{2}} \geqslant\mathrm{0} \\$$$${f}\left({x}\right)=\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}\left({b}+{c}+{d}+{e}+{f}\right)+\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{d}^{\mathrm{2}} +{e}^{\mathrm{2}} +{f}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\$$$${f}\left({x}\right)=\mathrm{5}{x}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{10}−{a}\right){x}+\left(\mathrm{25}−{a}^{\mathrm{2}} \right)\geqslant\mathrm{0} \\$$$$\Rightarrow\left(\mathrm{10}−{a}\right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{25}−{a}^{\mathrm{2}} \right)\leqslant\mathrm{0} \\$$$$\Rightarrow\mathrm{6}{a}^{\mathrm{2}} −\mathrm{20}{a}−\mathrm{25}\leqslant\mathrm{0} \\$$$$\Rightarrow\frac{\mathrm{10}−\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}}\leqslant{a}\leqslant\frac{\mathrm{10}+\mathrm{5}\sqrt{\mathrm{10}}}{\mathrm{6}} \\$$