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Math Help - Hard minimum value problem

  1. #1
    0-)
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    Hard minimum value problem

    Show that if x and y are real numbers and 7x^2+3xy+3y^2=1 then the least positive value of \frac{x^2+y^2}{y} is 0.5.

    I've tried lots of methods but none work.
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Ok

    [quote=0-);131057]Show that if x and y are real numbers and 7x^2+3xy+3y^2=1 then the least positive value of \frac{x^2+y^2}{y} is 0.5.

    Solve the first for either x or y...imput that into the second formula...differentiate and find the minimum...then to get the other value imput your one value into the original equation and solve for the other value...and to make sure you are getting a minimum plug your answer into the second derivative and make sure its positive
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  3. #3
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    [quote=Mathstud28;131089]
    Quote Originally Posted by 0-) View Post
    Show that if x and y are real numbers and 7x^2+3xy+3y^2=1 then the least positive value of \frac{x^2+y^2}{y} is 0.5.

    Solve the first for either x or y...imput that into the second formula...differentiate and find the minimum...then to get the other value imput your one value into the original equation and solve for the other value...and to make sure you are getting a minimum plug your answer into the second derivative and make sure its positive
    Thank you.

    I thought there would be a quicker way of doing it but your way should work.
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