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**ashura** Need help to continue with the following solution: Find the stationary points of the function, $\displaystyle f(x,y) = x^2 + y^2$

subject to the constraint, $\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0 $

$\displaystyle L_{xy \lambda} = x^2 + y^2 + \lambda (3x^2 + 7y^2 + 12x - 28y + 1)$

$\displaystyle L_x = 2x + \lambda (6x + 12)$

$\displaystyle L_y = 2y + \lambda (14y - 28)$

$\displaystyle L_ \lambda = 3x^2 + 7y^2 + 12x - 28y + 1$

$\displaystyle L_x = L_y = L_ \lambda = 0$

$\displaystyle 2x + \lambda (6x + 12) = 0$

$\displaystyle 2y + \lambda (14y - 28) = 0$

$\displaystyle 3x^2 + 7y^2 + 12x - 28y + 1 = 0$

$\displaystyle x + \lambda (3x + 6) = 0$

$\displaystyle y + \lambda (7y - 14) = 0$

$\displaystyle \lambda = \frac{-x}{3x + 6}$

$\displaystyle \lambda = - \frac{y}{7y - 14}$

equating, $\displaystyle - \frac{x}{3x + 6} = - \frac{y}{7y - 14}$

$\displaystyle x(7y -14) = y(3x + 6)$

$\displaystyle 7xy - 14x = 3xy + 6y$

$\displaystyle 7xy - 3xy - 14x = 6y$

$\displaystyle 4xy - 14x = 6y$

$\displaystyle x(4y - 14) = 6y$

$\displaystyle x = \frac {6y}{4y - 14}$

$\displaystyle 3 \left ( \frac{6y}{4y - 14} \right ) + 7y^2 + 12 \left ( \frac{6y}{4y - 14} \right )$