I am new with Lagrange multipliers and am having trouble starting, understanding, and knowing what steps to take using the method of Lagrange multipliers to find the minimum and maximum values of:
$$f(x,y) = x^2 y - 2/3y^3$$
such that $(x,y)$ lies on the graph of the ellipse whose equation is $3x^2+y^2 = 9.$
How will I know which is the constraint? Though I think it is $3x^2+y^2 = 9$ I am not sure, and if it is, how do I apply it to a partial derivative to get a new system of equations if it equates to $9?$ Do I just divide the whole equation with 9 then differentiate? Thank you for your help.
Putting the constraint in the form $c(x,y) = 0$, we have $c(x,y) = 3x^2 + y^2 - 9 $
Then define
$g(x,y) = f(x,y) + \lambda c(x,y) = x^2 y - \dfrac{2}{3} y^3 + \lambda (3x^2 + y^2 - 9) $
Take the derivatives of $g$ with respect to $x, y, \lambda$ and set them equal to zero.
$g_x = 2 x y + \lambda (6 x) = 0$
$g_y = x^2 + \lambda (2 y) = 0$
$g_\lambda = 3 x^2 + y^2 - 9 = 0$
By inspection, we can conclude that $(x, y)$ cannot be $(0,0)$
Hence, we have $ y = - 3 \lambda $, and $ x^2 = - 2 \lambda y = 6 \lambda^2 $
Substitute these into the third equation,
$ 18 \lambda^2 + 9 \lambda^2 - 9 = 0 $
From which $ \lambda = \pm \dfrac{1}{\sqrt{3}} $
For $\lambda = \dfrac{1}{\sqrt{3}}$, we get $ y = -\sqrt{3}, x = 2$
and therefore , $f(x,y) = -4 \sqrt{3} - \frac{2}{3} (-3\sqrt{3} ) = -2 \sqrt{3}$
And for $\lambda = - \dfrac{1}{\sqrt{3}}$ , we get $ y = \sqrt{3} , x = 2 $, so that $f(x, y) = 4 \sqrt{3} - \dfrac{2}{3} (3 \sqrt{3} ) = 2 \sqrt{3} $
Therefore the minimum of $(-2 \sqrt{3})$ occurs at $(2, -\sqrt{3})$ and the maximum of $2 \sqrt{3}$ occurs at $(2, \sqrt{3})$