Use Lagrange multiplier to find maximum and minimum of $f(x,y) = 3x-4y$ subject to $x^2+3y^2=129$.
So I start by getting the partial with respect to both f(x, y) and $x^2+3y^2=129$
Ultimately, I get
$f x(x, y) = 3$ $f y(x, y) = -4$
$g x(x, y) = 2x$ $g y(x, y) = 6y$
Then I equate them to get $3 = 2x(times constant)$ and $-4 = 6y(times constant)$
How do I simplify to get x and y. I tried solving for both the constant and the variables but wasn't getting a correct answer. Can anyone assist?
Best Answer
$$ [3,-4]=\lambda[2x,,6y]=[2x\lambda,6y\lambda] $$ $$ 3=2x\lambda, -4= 6y\lambda $$ $$ \lambda =3/(2x) = -4/(6y)= -2/(3y) $$ $$ 9y=-4x $$ Now substitute $ y=(-4/9)x $ into the constraint.