Question Number 79404 by jagoll last updated on 25/Jan/20
$$\mathrm{for}\:\mathrm{every}\:\mathrm{real}\:\mathrm{number}\:\mathrm{a}\:,\:\mathrm{b}\: \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{4a}−\mathrm{6b}=\mathrm{2}.\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\: \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{expression}\: \\ $$$$\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{8a}−\mathrm{10b}+\mathrm{41}}\:? \\ $$
Commented by john santu last updated on 25/Jan/20
$${let}\:{f}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{6}{b}−\left(\mathrm{4}{a}+\mathrm{4}{b}\right)+\mathrm{41}} \\ $$$${f}=\sqrt{\mathrm{43}−\left(\mathrm{4}{a}+\mathrm{4}{b}\right)}\:.\:{That}\:{the}\: \\ $$$${value}\:{of}\:{f}\:{will}\:{be}\:{maximum}\: \\ $$$${if}\:\:\mathrm{4}{a}+\mathrm{4}{b}\:{is}\:{minimum}. \\ $$$${let}\::\:\mathrm{4}{x}+\mathrm{4}{y}=\:{l}\:,\:{this}\:{is}\:{the}\:{tangent}\: \\ $$$${of}\:{circle}\:\left({a}−\mathrm{2}\right)^{\mathrm{2}} +\left({b}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{15} \\ $$$${d}=\:\frac{\mid\mathrm{8}+\mathrm{12}−{l}\mid}{\mathrm{4}\sqrt{\mathrm{2}}},\:{d}={r} \\ $$$$\mathrm{4}\sqrt{\mathrm{30}}\:=\:\mid\mathrm{20}−{l}\mid\: \\ $$$$\mathrm{20}−{l}\:=\:\pm\mathrm{4}\sqrt{\mathrm{30}}\:\Rightarrow{l}=\:\mathrm{20}\pm\mathrm{4}\sqrt{\mathrm{30}} \\ $$$$\left({i}\right)\:{l}=\mathrm{20}+\mathrm{4}\sqrt{\mathrm{30}}\: \\ $$$${f}=\sqrt{\mathrm{43}−\left(\mathrm{20}+\mathrm{4}\sqrt{\mathrm{30}}\right)}\:=\sqrt{\mathrm{23}−\mathrm{2}\sqrt{\mathrm{120}}}\: \\ $$$$=\:\sqrt{\mathrm{15}}−\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$$\left({ii}\right)\:{l}=\:\mathrm{20}−\mathrm{4}\sqrt{\mathrm{30}}\: \\ $$$${f}=\sqrt{\mathrm{43}−\left(\mathrm{20}−\mathrm{4}\sqrt{\mathrm{30}}\right)}\:=\sqrt{\mathrm{15}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\therefore\:{f}_{{max}} \:=\:\sqrt{\mathrm{15}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${f}_{{min}} =\sqrt{\mathrm{15}}\:−\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$
Commented by jagoll last updated on 25/Jan/20
$$\mathrm{great}!\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you} \\ $$
Commented by peter frank last updated on 25/Jan/20
$${great} \\ $$
Commented by john santu last updated on 25/Jan/20
$${thanks}\:{sir} \\ $$