Question-224763

Question Number 224763 by behi834171 last updated on 02/Oct/25

Commented by Ghisom_ last updated on 03/Oct/25

min f(x, y) is ≈3.67113410 at ((x),(y) ) = ((((5(√6)−3(√5))/7)),((3(√(10))−5(√3))) )

$$\mathrm{min}\:{f}\left({x},\:{y}\right)\:\mathrm{is}\:\approx\mathrm{3}.\mathrm{67113410}\:\mathrm{at} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\frac{\mathrm{5}\sqrt{\mathrm{6}}−\mathrm{3}\sqrt{\mathrm{5}}}{\mathrm{7}}}\\{\mathrm{3}\sqrt{\mathrm{10}}−\mathrm{5}\sqrt{\mathrm{3}}}\end{pmatrix} \\ $$

Commented by behi834171 last updated on 04/Oct/25

$${thank}\:{you}\:{very}\:{much}\:{sir}. \\ $$$${please}\:{send}\:{your}\:{method},{if}\:{possible}. \\ $$

Commented by Ghisom_ last updated on 04/Oct/25

(∂f/∂x)=0 ⇒ solve for x (∂f/∂y)=0 ⇒ solve for y insert y (or) x and solve I will post the complete path later

$$\frac{\partial{f}}{\partial{x}}=\mathrm{0}\:\Rightarrow\:\mathrm{solve}\:\mathrm{for}\:{x} \\ $$$$\frac{\partial{f}}{\partial{y}}=\mathrm{0}\:\Rightarrow\:\mathrm{solve}\:\mathrm{for}\:{y} \\ $$$$\mathrm{insert}\:{y}\:\left(\mathrm{or}\right)\:{x}\:\mathrm{and}\:\mathrm{solve} \\ $$$$\mathrm{I}\:\mathrm{will}\:\mathrm{post}\:\mathrm{the}\:\mathrm{complete}\:\mathrm{path}\:\mathrm{later} \\ $$

Answered by mr W last updated on 06/Oct/25

Commented by mr W last updated on 06/Oct/25

i′ve developed following geometrical solution. with a=(√3), b=(√5) f(x,y)=(√(x^2 −(√2)ax+a^2 ))+(√(x^2 +y^2 −(√2)xy))+(√(y^2 −(√2)by+b^2 )) =(√(x^2 −2ax cos 45°+a^2 ))+(√(x^2 +y^2 −2xy cos 45°))+(√(y^2 −2by cos 45°+b^2 )) =AC+CD+DB f(x,y)_(min) is the shortest path from A to B. f(x,y)_(min) =AB=d=(√(a^2 +b^2 −2ab cos 135°)) =(√(a^2 +b^2 +(√2)ab)) =(√(3+5+(√(2×3×5))))=(√(8+(√(30))))≈3.671134 ((sin α)/b)=((sin β)/a)=((sin 135°)/d)=(1/( (√2)d)) ⇒sin α=(b/( (√2)d)), sin β=(a/( (√2)d)) (x/(sin α))=(a/(sin (90°+β))) ⇒x=((a sin α)/(cos β))=((ab)/( (√2)d×(√(1−((a/( (√2)d)))^2 )))) =((ab)/( (√(2d^2 −a^2 ))))=((ab)/( (√(a^2 +2b^2 +2(√2)ab)))) =((ab)/(a+(√2)b)) =((√(3×5))/( (√3)+(√(10))))=((5(√6)−3(√5))/( 7)) similarly y=((ab)/( (√2)a+b)) =((√(3×5))/( (√6)+(√5)))=3(√(10))−5(√3)

$${i}'{ve}\:{developed}\:{following}\:{geometrical} \\ $$$${solution}. \\ $$$${with}\:{a}=\sqrt{\mathrm{3}},\:{b}=\sqrt{\mathrm{5}} \\ $$$${f}\left({x},{y}\right)=\sqrt{{x}^{\mathrm{2}} −\sqrt{\mathrm{2}}{ax}+{a}^{\mathrm{2}} }+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\sqrt{\mathrm{2}}{xy}}+\sqrt{{y}^{\mathrm{2}} −\sqrt{\mathrm{2}}{by}+{b}^{\mathrm{2}} } \\ $$$$\:\:\:\:=\sqrt{{x}^{\mathrm{2}} −\mathrm{2}{ax}\:\mathrm{cos}\:\mathrm{45}°+{a}^{\mathrm{2}} }+\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}\:\mathrm{cos}\:\mathrm{45}°}+\sqrt{{y}^{\mathrm{2}} −\mathrm{2}{by}\:\mathrm{cos}\:\mathrm{45}°+{b}^{\mathrm{2}} } \\ $$$$\:\:\:\:={AC}+{CD}+{DB} \\ $$$${f}\left({x},{y}\right)_{{min}} \:{is}\:{the}\:{shortest}\:{path}\:{from}\:{A}\:{to}\:{B}. \\ $$$${f}\left({x},{y}\right)_{{min}} ={AB}={d}=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} −\mathrm{2}{ab}\:\mathrm{cos}\:\mathrm{135}°} \\ $$$$\:\:\:\:\:\:=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{2}}{ab}} \\ $$$$\:\:\:\:\:\:=\sqrt{\mathrm{3}+\mathrm{5}+\sqrt{\mathrm{2}×\mathrm{3}×\mathrm{5}}}=\sqrt{\mathrm{8}+\sqrt{\mathrm{30}}}\approx\mathrm{3}.\mathrm{671134} \\ $$$$ \\ $$$$\frac{\mathrm{sin}\:\alpha}{{b}}=\frac{\mathrm{sin}\:\beta}{{a}}=\frac{\mathrm{sin}\:\mathrm{135}°}{{d}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}{d}} \\ $$$$\Rightarrow\mathrm{sin}\:\alpha=\frac{{b}}{\:\sqrt{\mathrm{2}}{d}},\:\mathrm{sin}\:\beta=\frac{{a}}{\:\sqrt{\mathrm{2}}{d}} \\ $$$$\frac{{x}}{\mathrm{sin}\:\alpha}=\frac{{a}}{\mathrm{sin}\:\left(\mathrm{90}°+\beta\right)} \\ $$$$\Rightarrow{x}=\frac{{a}\:\mathrm{sin}\:\alpha}{\mathrm{cos}\:\beta}=\frac{{ab}}{\:\sqrt{\mathrm{2}}{d}×\sqrt{\mathrm{1}−\left(\frac{{a}}{\:\sqrt{\mathrm{2}}{d}}\right)^{\mathrm{2}} }} \\ $$$$\:\:\:\:\:\:\:=\frac{{ab}}{\:\sqrt{\mathrm{2}{d}^{\mathrm{2}} −{a}^{\mathrm{2}} }}=\frac{{ab}}{\:\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{ab}}} \\ $$$$\:\:\:\:\:\:\:=\frac{{ab}}{{a}+\sqrt{\mathrm{2}}{b}} \\ $$$$\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{3}×\mathrm{5}}}{\:\sqrt{\mathrm{3}}+\sqrt{\mathrm{10}}}=\frac{\mathrm{5}\sqrt{\mathrm{6}}−\mathrm{3}\sqrt{\mathrm{5}}}{\:\mathrm{7}} \\ $$$${similarly} \\ $$$${y}=\frac{{ab}}{\:\sqrt{\mathrm{2}}{a}+{b}} \\ $$$$\:\:\:=\frac{\sqrt{\mathrm{3}×\mathrm{5}}}{\:\sqrt{\mathrm{6}}+\sqrt{\mathrm{5}}}=\mathrm{3}\sqrt{\mathrm{10}}−\mathrm{5}\sqrt{\mathrm{3}} \\ $$

Commented by behi834171 last updated on 06/Oct/25