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 $0$?

This is from your working -
$(3x^2 -3, 3y^2 -3) = \lambda (1,2)$
$3x^2 - 3 = \lambda, 3y^2-3 = 2\lambda$
Equating $\lambda$ from both equations,
$6x^2-6 = 3y^2-3 \implies 2x^2 - y^2 = 1$
Substitute $x$ from $x+2y = 3$
$2(3-2y)^2 - y^2 = 1$
$\implies 7y^2 - 24y + 17 = 0 \, $ or $(7y-17)(y-1) = 0$
Can you take it from here and find possible points for extrema?
