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Question Number 218153 by Rojarani last updated on 31/Mar/25
x+y =12 minimum value of (√(x^2 +4)) +(√(y^2 +9)) =?
$$\:{x}+{y}\:=\mathrm{12} \\ $$$$\:{minimum}\:{value}\:{of} \\ $$$$\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}\:=? \\ $$
Answered by mr W last updated on 31/Mar/25
Commented by Ghisom last updated on 31/Mar/25
nice!
$$\mathrm{nice}! \\ $$
Commented by mr W last updated on 31/Mar/25
AC=(√(x^2 +2^2 ))=(√(x^2 +4)) BC=(√(y^2 +3^2 ))=(√(y^2 +9)) AB=(√(12^2 +(2+3)^2 ))=13 AC+BC≥AB ⇒(√(x^2 +4))+(√(y^2 +9))≥13 i.e. ((√(x^2 +4))+(√(y^2 +9)))_(min) =13
$${AC}=\sqrt{{x}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\sqrt{{x}^{\mathrm{2}} +\mathrm{4}} \\ $$$${BC}=\sqrt{{y}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\sqrt{{y}^{\mathrm{2}} +\mathrm{9}} \\ $$$${AB}=\sqrt{\mathrm{12}^{\mathrm{2}} +\left(\mathrm{2}+\mathrm{3}\right)^{\mathrm{2}} }=\mathrm{13} \\ $$$${AC}+{BC}\geqslant{AB} \\ $$$$\Rightarrow\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}\geqslant\mathrm{13} \\ $$$${i}.{e}.\:\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{y}^{\mathrm{2}} +\mathrm{9}}\right)_{{min}} =\mathrm{13} \\ $$
Commented by Rojarani last updated on 31/Mar/25
Sir, thanks.
$$\:{Sir},\:{thanks}. \\ $$
Commented by mehdee7396 last updated on 01/Apr/25
very good ⋛
$${very}\:\:{good}\:\:\:\underline{\underbrace{\lesseqgtr}} \\ $$
Commented by mr W last updated on 01/Apr/25
thanks to all!
$${thanks}\:{to}\:{all}! \\ $$
Answered by efronzo1 last updated on 31/Mar/25
$$\:\cancel{\:} \\ $$
Answered by vnm last updated on 31/Mar/25
x=6+t, y=6−t (d/dt)((√((6+t)^2 +4))+(√((6−t)^2 +9)))= ((6+t)/( (√((6+t)^2 +4))))+((6−t)/( (√((6−t)^2 +9))))= (((6+t)(√((6−t)^2 +9))+(6−t)(√((6+t)^2 +4)))/( (√((6+t)^2 +4))(√((6−t)^2 +9))))=0 (t+6)(√(t^2 −12t+45))=(t−6)(√(t^2 +12t+40)) (t^2 +12t+36)(t^2 −12t+45)= (t^2 −12t+36)(t^2 +12t+40) (t^2 +12t)45+36(−12t+45)= (t^2 −12t)40+36(12t+40) 5t^2 +12t∙85−36∙24t+36∙5=0 5t^2 +156t+180=0 t=((−78±72)/5) t_1 =−30 in this point (d/dt)(x^2 +4)>0 and (d/dt)(y^2 +9)>0, therefor this point isn′t the point of minimum t_2 =−(6/5) (√((6−(6/5))^2 +4))+(√((6+(6/5))^2 +9))=13
$${x}=\mathrm{6}+{t},\:\:\:{y}=\mathrm{6}−{t} \\ $$$$\frac{{d}}{{dt}}\left(\sqrt{\left(\mathrm{6}+{t}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\mathrm{6}−{t}\right)^{\mathrm{2}} +\mathrm{9}}\right)= \\ $$$$\frac{\mathrm{6}+{t}}{\:\sqrt{\left(\mathrm{6}+{t}\right)^{\mathrm{2}} +\mathrm{4}}}+\frac{\mathrm{6}−{t}}{\:\sqrt{\left(\mathrm{6}−{t}\right)^{\mathrm{2}} +\mathrm{9}}}= \\ $$$$\frac{\left(\mathrm{6}+{t}\right)\sqrt{\left(\mathrm{6}−{t}\right)^{\mathrm{2}} +\mathrm{9}}+\left(\mathrm{6}−{t}\right)\sqrt{\left(\mathrm{6}+{t}\right)^{\mathrm{2}} +\mathrm{4}}}{\:\sqrt{\left(\mathrm{6}+{t}\right)^{\mathrm{2}} +\mathrm{4}}\sqrt{\left(\mathrm{6}−{t}\right)^{\mathrm{2}} +\mathrm{9}}}=\mathrm{0} \\ $$$$\left({t}+\mathrm{6}\right)\sqrt{{t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{45}}=\left({t}−\mathrm{6}\right)\sqrt{{t}^{\mathrm{2}} +\mathrm{12}{t}+\mathrm{40}} \\ $$$$\left({t}^{\mathrm{2}} +\mathrm{12}{t}+\mathrm{36}\right)\left({t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{45}\right)= \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{12}{t}+\mathrm{36}\right)\left({t}^{\mathrm{2}} +\mathrm{12}{t}+\mathrm{40}\right) \\ $$$$\left({t}^{\mathrm{2}} +\mathrm{12}{t}\right)\mathrm{45}+\mathrm{36}\left(−\mathrm{12}{t}+\mathrm{45}\right)= \\ $$$$\left({t}^{\mathrm{2}} −\mathrm{12}{t}\right)\mathrm{40}+\mathrm{36}\left(\mathrm{12}{t}+\mathrm{40}\right) \\ $$$$\mathrm{5}{t}^{\mathrm{2}} +\mathrm{12}{t}\centerdot\mathrm{85}−\mathrm{36}\centerdot\mathrm{24}{t}+\mathrm{36}\centerdot\mathrm{5}=\mathrm{0} \\ $$$$\mathrm{5}{t}^{\mathrm{2}} +\mathrm{156}{t}+\mathrm{180}=\mathrm{0} \\ $$$${t}=\frac{−\mathrm{78}\pm\mathrm{72}}{\mathrm{5}} \\ $$$${t}_{\mathrm{1}} =−\mathrm{30}\:{in}\:{this}\:{point}\:\frac{{d}}{{dt}}\left({x}^{\mathrm{2}} +\mathrm{4}\right)>\mathrm{0}\:{and}\:\frac{{d}}{{dt}}\left({y}^{\mathrm{2}} +\mathrm{9}\right)>\mathrm{0},\: \\ $$$${therefor}\:{this}\:{point}\:{isn}'{t}\:{the}\:{point}\:{of}\:{minimum} \\ $$$${t}_{\mathrm{2}} =−\frac{\mathrm{6}}{\mathrm{5}} \\ $$$$\sqrt{\left(\mathrm{6}−\frac{\mathrm{6}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\mathrm{6}+\frac{\mathrm{6}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{9}}=\mathrm{13} \\ $$$$ \\ $$

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