Respuesta :

ghanami
hello :
let : M(x,y) in the curve  and  the origin : O (0,0)
calculate : OA
OA² = x²+y².....(1)
but : xy² = 250
so : y² = 250/x
subsct in (1) :
OA² =  x² + 250/x
 let : f(x) = x² +250/x
calculate the minumum of : f
f'(x) = 2x - 250/x²
f'(x) = 0 : 250/x² = 2x
 x^3 = 125 = 5^3
x= 5
f(5) =5² +250/3 =325/3....(The closest distance)

M(x,y) in the curve  : x=5 : 5y² = 250
y² =50
y = √(50) or y = - √(50)
two points on the curve xy 2 = 250 that are closest to the origin:
M1 (5 , √(50))   , M2 (5 , -√(50)) 
Ver imagen ghanami
caylus
Hello,
All we have to do it to minimise [tex]f= \sqrt{x^2+y^2} = \sqrt{x^2+ \dfrac{250}{x^2} } \\ \dfrac{df}{dx} = \dfrac{1}{2}* \dfrac{2x- \dfrac{2*250}{x^3}}{ \sqrt{x^2+ \dfrac{250}{x^2} } }\\ \rightarrow x- \dfrac{250}{x^3} =0\\ \rightarrow x= \sqrt[4]{250} \approx 3.9763\\ \rightarrow y=250^{ \frac{3}{8} }\approx 7,9292\\ Sol=\{(\sqrt[4]{250}, \sqrt[8]{250^3} ), (\sqrt[4]{250}, -\sqrt[8]{250^3} )\} \\ [/tex]

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