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Question Number 140417 by mathsuji last updated on 07/May/21
if minimum value of g(a;b)=(√(a^2 +b^2 −10a−10b+50))+(√(b^2 −4y+20))+ +(√(a^2 −14a+74)) is n and occurs at a=γ , b=δ, the find (n+4γ+3δ)=?
$${if}\:{minimum}\:{value}\:{of} \\ $$$${g}\left({a};{b}\right)=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{10}{a}−\mathrm{10}{b}+\mathrm{50}}+\sqrt{{b}^{\mathrm{2}} −\mathrm{4}{y}+\mathrm{20}}+ \\ $$$$+\sqrt{{a}^{\mathrm{2}} −\mathrm{14}{a}+\mathrm{74}} \\ $$$${is}\:{n}\:{and}\:{occurs}\:{at}\:{a}=\gamma\:,\:{b}=\delta,\:{the}\:{find} \\ $$$$\left({n}+\mathrm{4}\gamma+\mathrm{3}\delta\right)=? \\ $$
Answered by mr W last updated on 07/May/21
g(a,b)=(√((a−5)^2 +(b−5)^2 ))+(√((b−2)^2 +16))+(√((a−7)^2 +25)) =(√((a−5)^2 +(5−b)^2 ))+(√(4^2 +(b−2)^2 ))+(√((7−a)^2 +5^2 )) ≥(√((a−5+4+7−a)^2 +(5−b+b−2+5)^2 )) =(√(6^2 +8^2 )) =10 minimum n=10 occurs when ((a−5)/(5−b))=(4/(b−2))=((7−a)/5)=(6/8) ⇒a=7−((5×6)/8)=((13)/4)=γ ⇒b=2+((4×8)/6)=((22)/3)=δ ⇒n+4γ+3δ=10+13+22=45
$${g}\left({a},{b}\right)=\sqrt{\left({a}−\mathrm{5}\right)^{\mathrm{2}} +\left({b}−\mathrm{5}\right)^{\mathrm{2}} }+\sqrt{\left({b}−\mathrm{2}\right)^{\mathrm{2}} +\mathrm{16}}+\sqrt{\left({a}−\mathrm{7}\right)^{\mathrm{2}} +\mathrm{25}} \\ $$$$=\sqrt{\left({a}−\mathrm{5}\right)^{\mathrm{2}} +\left(\mathrm{5}−{b}\right)^{\mathrm{2}} }+\sqrt{\mathrm{4}^{\mathrm{2}} +\left({b}−\mathrm{2}\right)^{\mathrm{2}} }+\sqrt{\left(\mathrm{7}−{a}\right)^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} } \\ $$$$\geqslant\sqrt{\left({a}−\mathrm{5}+\mathrm{4}+\mathrm{7}−{a}\right)^{\mathrm{2}} +\left(\mathrm{5}−{b}+{b}−\mathrm{2}+\mathrm{5}\right)^{\mathrm{2}} } \\ $$$$=\sqrt{\mathrm{6}^{\mathrm{2}} +\mathrm{8}^{\mathrm{2}} } \\ $$$$=\mathrm{10} \\ $$$${minimum}\:{n}=\mathrm{10}\:{occurs}\:{when} \\ $$$$\frac{{a}−\mathrm{5}}{\mathrm{5}−{b}}=\frac{\mathrm{4}}{{b}−\mathrm{2}}=\frac{\mathrm{7}−{a}}{\mathrm{5}}=\frac{\mathrm{6}}{\mathrm{8}} \\ $$$$\Rightarrow{a}=\mathrm{7}−\frac{\mathrm{5}×\mathrm{6}}{\mathrm{8}}=\frac{\mathrm{13}}{\mathrm{4}}=\gamma \\ $$$$\Rightarrow{b}=\mathrm{2}+\frac{\mathrm{4}×\mathrm{8}}{\mathrm{6}}=\frac{\mathrm{22}}{\mathrm{3}}=\delta \\ $$$$\Rightarrow{n}+\mathrm{4}\gamma+\mathrm{3}\delta=\mathrm{10}+\mathrm{13}+\mathrm{22}=\mathrm{45} \\ $$
Commented by mr W last updated on 07/May/21
minkowski inequality was applied: (√(a^2 +b^2 ))+(√(c^2 +d^2 ))+(√(e^2 +f^2 ))≥(√((a+c+e)^2 +(b+d+f)^2 ))
$${minkowski}\:{inequality}\:{was}\:{applied}: \\ $$$$\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{c}^{\mathrm{2}} +{d}^{\mathrm{2}} }+\sqrt{{e}^{\mathrm{2}} +{f}^{\mathrm{2}} }\geqslant\sqrt{\left({a}+{c}+{e}\right)^{\mathrm{2}} +\left({b}+{d}+{f}\right)^{\mathrm{2}} } \\ $$
Commented by mathsuji last updated on 09/May/21
perfect thankyou Sir
$${perfect}\:{thankyou}\:{Sir} \\ $$

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