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Math Help - the last part

  1. #1
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    the last part

    find the extremea of f(x,y) = x^2 -4y^2 subject to g(x,y) = x^2+ 4xy +6y^2 =2o.

    making use of lagrange multiplier, i got

    f(x,y) has a min or max when x^2 +5xy+4y^2 = o and that it has to also fulfil the conditions of g(x,y) = x^2+ 4xy +6y^2 =2o right?

    when i solve these 2 equations via simultaneous, i got y=0 and x=2y. then how do i solve the last part?
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  2. #2
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    Quote Originally Posted by alexandrabel90 View Post
    find the extremea of f(x,y) = x^2 -4y^2 subject to g(x,y) = x^2+ 4xy +6y^2 =2o.

    making use of lagrange multiplier, i got

    f(x,y) has a min or max when x^2 +5xy+4y^2 = o and that it has to also fulfil the conditions of g(x,y) = x^2+ 4xy +6y^2 =2o right?

    when i solve these 2 equations via simultaneous, i got y=0 and x=2y. then how do i solve the last part?
    It took me a moment to see how you got x^2+ 5xy+ 4y^2= 0 but, yes, that is correct. Subtracting that from x^2+ 4xy+ 6y^2= 20 gives -xy+ 2y^2= 20. That does NOT give "y= 0 and x= 2y". They satisfy y(x- 2y)= y^2- 2y= 0. Did you forget the "20"?

    You can solve -xy+ 2y^2= 20 for x as x= \frac{2y^2- 20}{y}. Put that back into, say, x^2+ 4xy+ 6y^2= 20 to get an equation for y alone.
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  3. #3
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    subbing x into x^2+ 4xy +6y^2 =2o , i got 10y^4 -172y^2 +400=0.

    so y^2 = (172 + (29544)^(1/2) )/ 20 or y^2 = (172 - (29544)^(1/2) )/ 20

    then how do i get an equation in terms of just y alone and not y^2?


    from here, what do i sub into x and y to find the points where f has a local minima and maxima?

    thanks!
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