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 $0$?
 A: 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.
A: 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?
A: In this case, it is really not a great idea to use Lagrange multipliers. We can write $x$ in terms of $y$ (or vice-versa) using the restriction and reduce this question to a one variable optimisation problem. Substituting $x = 3 - 2y$, the problem reduces to finding the extrema of $f(y)=-7 y^3+36 y^2-51 y+18$. The critical points are $y = \frac{17}{7}$ and $y = 1$ and, considering that $f''(1)>0$ and $f''(\frac{17}{7})<0$, they are a local minimum and maximum, respectively.
It is also worth noticing that, since $\displaystyle \lim_{y \to \pm \infty} f(y) = \mp \infty$, there will not be any global extrema.
