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Question Number 80027 by jagoll last updated on 30/Jan/20

find minimum value of (√(x^2 +4))+(√(x^2 −24x+153)) for x≥0 in R

$${find}\:{minimum} \\ $$$${value}\:{of}\:\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}+\sqrt{{x}^{\mathrm{2}} −\mathrm{24}{x}+\mathrm{153}} \\ $$$${for}\:{x}\geqslant\mathrm{0}\:{in}\:\mathbb{R} \\ $$

Commented by john santu last updated on 30/Jan/20

yes sir

$$\mathrm{yes}\:\mathrm{sir} \\ $$

Commented by john santu last updated on 30/Jan/20

f′(x) = (x/(√(x^2 +4)))+((x−12)/(√(x^2 −24x+153))) =0 (x/(12−x)) =(√((x^2 +4)/(x^2 −24x+153))) (x^2 /(144−24x+x^2 )) =((x^2 +x)/(x^2 −24x+153)) x^4 −24x^3 +153x^2 =x^4 −24x^3 +144x^2 +4x^2 −96x+576 5x^2 +96x−576=0 x =((−96+(√(20736)))/(10)) = ((−96+144)/(10))=4.8 f_(min) = (√((((24)/5))^2 +4))+(√((((24)/5)−12)^2 +9))

$$\mathrm{f}'\left(\mathrm{x}\right)\:=\:\frac{\mathrm{x}}{\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}}+\frac{\mathrm{x}−\mathrm{12}}{\sqrt{\mathrm{x}^{\mathrm{2}} −\mathrm{24x}+\mathrm{153}}}\:=\mathrm{0} \\ $$$$\frac{\mathrm{x}}{\mathrm{12}−\mathrm{x}}\:=\sqrt{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{4}}{\mathrm{x}^{\mathrm{2}} −\mathrm{24x}+\mathrm{153}}} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{144}−\mathrm{24x}+\mathrm{x}^{\mathrm{2}} }\:=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{x}}{\mathrm{x}^{\mathrm{2}} −\mathrm{24x}+\mathrm{153}} \\ $$$$\mathrm{x}^{\mathrm{4}} −\mathrm{24x}^{\mathrm{3}} +\mathrm{153x}^{\mathrm{2}} =\mathrm{x}^{\mathrm{4}} −\mathrm{24x}^{\mathrm{3}} +\mathrm{144x}^{\mathrm{2}} \\ $$$$+\mathrm{4x}^{\mathrm{2}} −\mathrm{96x}+\mathrm{576} \\ $$$$\mathrm{5x}^{\mathrm{2}} +\mathrm{96x}−\mathrm{576}=\mathrm{0} \\ $$$$\mathrm{x}\:=\frac{−\mathrm{96}+\sqrt{\mathrm{20736}}}{\mathrm{10}}\:=\:\frac{−\mathrm{96}+\mathrm{144}}{\mathrm{10}}=\mathrm{4}.\mathrm{8} \\ $$$$\mathrm{f}_{\mathrm{min}} =\:\sqrt{\left(\frac{\mathrm{24}}{\mathrm{5}}\right)^{\mathrm{2}} +\mathrm{4}}+\sqrt{\left(\frac{\mathrm{24}}{\mathrm{5}}−\mathrm{12}\right)^{\mathrm{2}} +\mathrm{9}} \\ $$$$\:\:\:\:\:\:\: \\ $$

Commented by mr W last updated on 30/Jan/20

with x=4.8 we get exactly f=13. i.e. f_(min) =13.

$${with}\:{x}=\mathrm{4}.\mathrm{8}\:{we}\:{get}\:{exactly}\:{f}=\mathrm{13}. \\ $$$${i}.{e}.\:{f}_{{min}} =\mathrm{13}. \\ $$

Commented by jagoll last updated on 30/Jan/20

thanks mr w and john

$${thanks}\:{mr}\:{w}\:{and}\:{john} \\ $$

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