The problem states: Use Lagrange multipliers to find the absolute minimum and maximum values of the function $f(x,y)=x^2y$ subject to the constraint $3x^2 + 2y^2 = 4$.
My attempt at a solution:
I am using Lagrange multipliers and obtain these equations. These are the equations.
$$f_x=\lambda g_x$$ $$f_y=\lambda g_y $$ $$g(x,y)=4$$ $$f_x=2xy$$ $$g_x=6x$$
$$f_y = x^2$$ $$g_y = 4y$$ $$g(x,y)=3x^2+2y^2$$
Plugging in the equations and coming up with a system of three equations and three unknowns yields.
$$2xy=\lambda 6x$$ $$x^2=\lambda 4y$$ $$3x^2+2y^2=4$$
I cannot figure out how to take it further to find the absolute maximum and minimums.
Best Answer
The equations you arrived at are:
$$ 2xy = \lambda 6 x$$
And,
$$ x^2 = \lambda 4y$$
Considering the first equation:
$$ x[ 2y - 6 \lambda ] = 0$$
So, either $ y= 3 \lambda$ or $ x=0$
According to the second equation, if $ x=0$ then $ y=0$ but this doesn't satisfy the constraint so let's throw this solution away. Now if $ y= 3 \lambda$ then, $ x^2 = 12 \lambda^2$ and plugging this and $ y= 3 \lambda$ into constraint gives:
$$ 3( 12 \lambda^2) + 2(3 \lambda)^2=4$$
$$ 36 \lambda^2 + 18 \lambda^2 -4 =0$$
And,
$$ y= 3 \lambda$$
$$ x^2 = 3 \lambda^2$$
Now, you have two $\lambda$ from the quadratic and each corresponds to an $ x-y$ minima/maxima. To check which if maxima or minima, you'll have to plug it into the function. ( I think you could use hessian here to identify maxima and minima but I don't understand that concept enough to explain).
Tl;dr : Plug result of lagrange multipliers into the constraint equation