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Question Number 79404 by jagoll last updated on 25/Jan/20

for every real number a , b such that a^2 +b^2 −4a−6b=2. what is the maximum and minimum value of the expression (√(a^2 +b^2 −8a−10b+41)) ?

$$\mathrm{for}\:\mathrm{every}\:\mathrm{real}\:\mathrm{number}\:\mathrm{a}\:,\:\mathrm{b}\: \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{4a}−\mathrm{6b}=\mathrm{2}.\: \\ $$$$\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{and}\: \\ $$$$\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{expression}\: \\ $$$$\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} −\mathrm{8a}−\mathrm{10b}+\mathrm{41}}\:? \\ $$

Commented by john santu last updated on 25/Jan/20

let f=(√(a^2 +b^2 −4a−6b−(4a+4b)+41)) f=(√(43−(4a+4b))) . That the value of f will be maximum if 4a+4b is minimum. let : 4x+4y= l , this is the tangent of circle (a−2)^2 +(b−3)^2 =15 d= ((∣8+12−l∣)/(4(√2))), d=r 4(√(30)) = ∣20−l∣ 20−l = ±4(√(30)) ⇒l= 20±4(√(30)) (i) l=20+4(√(30)) f=(√(43−(20+4(√(30))))) =(√(23−2(√(120)))) = (√(15))−2(√2) (ii) l= 20−4(√(30)) f=(√(43−(20−4(√(30))))) =(√(15))+2(√2) ∴ f_(max) = (√(15))+2(√2) f_(min) =(√(15)) −2(√2)

$${let}\:{f}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{4}{a}−\mathrm{6}{b}−\left(\mathrm{4}{a}+\mathrm{4}{b}\right)+\mathrm{41}} \\ $$$${f}=\sqrt{\mathrm{43}−\left(\mathrm{4}{a}+\mathrm{4}{b}\right)}\:.\:{That}\:{the}\: \\ $$$${value}\:{of}\:{f}\:{will}\:{be}\:{maximum}\: \\ $$$${if}\:\:\mathrm{4}{a}+\mathrm{4}{b}\:{is}\:{minimum}. \\ $$$${let}\::\:\mathrm{4}{x}+\mathrm{4}{y}=\:{l}\:,\:{this}\:{is}\:{the}\:{tangent}\: \\ $$$${of}\:{circle}\:\left({a}−\mathrm{2}\right)^{\mathrm{2}} +\left({b}−\mathrm{3}\right)^{\mathrm{2}} =\mathrm{15} \\ $$$${d}=\:\frac{\mid\mathrm{8}+\mathrm{12}−{l}\mid}{\mathrm{4}\sqrt{\mathrm{2}}},\:{d}={r} \\ $$$$\mathrm{4}\sqrt{\mathrm{30}}\:=\:\mid\mathrm{20}−{l}\mid\: \\ $$$$\mathrm{20}−{l}\:=\:\pm\mathrm{4}\sqrt{\mathrm{30}}\:\Rightarrow{l}=\:\mathrm{20}\pm\mathrm{4}\sqrt{\mathrm{30}} \\ $$$$\left({i}\right)\:{l}=\mathrm{20}+\mathrm{4}\sqrt{\mathrm{30}}\: \\ $$$${f}=\sqrt{\mathrm{43}−\left(\mathrm{20}+\mathrm{4}\sqrt{\mathrm{30}}\right)}\:=\sqrt{\mathrm{23}−\mathrm{2}\sqrt{\mathrm{120}}}\: \\ $$$$=\:\sqrt{\mathrm{15}}−\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$$$\left({ii}\right)\:{l}=\:\mathrm{20}−\mathrm{4}\sqrt{\mathrm{30}}\: \\ $$$${f}=\sqrt{\mathrm{43}−\left(\mathrm{20}−\mathrm{4}\sqrt{\mathrm{30}}\right)}\:=\sqrt{\mathrm{15}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\therefore\:{f}_{{max}} \:=\:\sqrt{\mathrm{15}}+\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${f}_{{min}} =\sqrt{\mathrm{15}}\:−\mathrm{2}\sqrt{\mathrm{2}}\: \\ $$

Commented by jagoll last updated on 25/Jan/20

great! sir thank you

$$\mathrm{great}!\:\mathrm{sir}\:\mathrm{thank}\:\mathrm{you} \\ $$

Commented by peter frank last updated on 25/Jan/20

great

$${great} \\ $$

Commented by john santu last updated on 25/Jan/20

thanks sir

$${thanks}\:{sir} \\ $$

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