Find maxima and minima of $f(x,y) = x^3 + y^3 -3x -3y$ in $x + 2y=3$ I need to find max and min of $f(x,y)=x^3 + y^3 -3x -3y$ with the following restriction: $x + 2y = 3$.
I used the multiplier's Lagrange theorem and found $(1,1)$ is the minima of $f$. Apparently, the maxima is $(-13/7, 17/7)$ but I could not find it via Lagrange's theorem.
Here's what I did:
I put up the linear system:
$\nabla f(x,y) = \lambda \, \nabla g(x,y)$
$g(x,y) = 0$
then,
$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$
$x + 2y -3 = 0$
Solving for $\lambda$, I got $\lambda = 0$, which gave me $x = 1$ and $y = 1$.
How can I find the maxima if lambda only gives one value which is $0$?

As others have said, you don't necessarily need to use Lagrange multipliers. But since you've set up the system, we can see what happens:
$$
3x^2-3=\lambda\\
3y^2-3=2\lambda\\
x+2y-3=0
$$
From the first two equations, we have $3y^2-3=2(3x^2-3)$, which simplifies to $y^2-1=2(x^2-1)$. Rearranging the linear constraint, we have $x=3-2y$. Putting this information together leads to
$$7y^2-24y+17=0.$$
You could solve this using the quadratic formula, but it is quicker to observe that $-24 = -7-17$:
$$7y^2-7y-17y+17=0$$
and so $(7y-17)(y-1)=0$.
This will give you the two local extrema.
