Find the max and min values of a multivariable function on the boundary of a domain I have the multivariable function: $f(x,y) = x^3 + y^3 - 3xy + 2$
I want to find the maximum and minimum values of this function on the domain: $$ D=[(x,y) : x,y\ge 0, x^2 +y^2\le4] $$
I found the partial derivatives to be:
$f_x=3x^2 -3y$
$f_y=3y^2-3x$
From here I set both my partial derivatives to equal $0$ and solved for $x$ and $y$ and got $(1,1)$ and $(0,0)$, I believe these two could potentially be the max/min I'm looking for, however I know there are more possibilities.
I then created a function $g$ where $g=x^2 + y^2$ ($g = 4$)
I know that $\nabla f= \lambda \nabla g$ so $(3x^2-3y,3y^2-3x) = \lambda(2x,2y) $
I then set up $3$ equations:
$\lambda x = \frac32  (x^2) - \frac32  (y)$
$\lambda y = \frac32  (y^2) - \frac32  (x)$
$x^2 + y^2 = 4$
But I'm not sure where to go from here and how to solve these equations for $x$ and $y$ and $\lambda$. To be honest I'm not even sure if this is the best way to go about answering this question.
If anyone could show me how I would find the maximum and minimum of the function on this domain with a better method, or help me continue with mine, it would really help.
 A: In both methods, you will have to additionally check boundary points for extrema.
By the way, Lagrange Multiplier method in this case does not directly give you the points where extrema occurs.
Applying Lagrange Multiplier method,
$3x^2 - 3 y = 2 \lambda x \tag1$
$3y^2 - 3 x = 2 \lambda y \tag2$
$x^2+y^2 \leq 4 \tag3$
If $x, y \ne 0$, from $(1), \lambda = \cfrac{3x^2-3y}{2x}$
Plugging into $(2), \cfrac{3y^2-3x}{2y} = \cfrac{3x^2-3y}{2x}$
$xy^2 - x^2 = x^2y - y^2$
$(x-y) (x+y+xy) = 0$
So $x = y$ is a solution.
Plugging into objective function, we have
$x^3 + y^3 - 3xy + 2 = 2x^3 - 3x^2 + 2 = (x-1)^2 (2x + 1) + 1$
As $x \geq 0$, the minimum occurs when $x = y = 1$
When $x^2 + y^2 = 4$, $x = y$ gives point $(\sqrt2, \sqrt2)$ that you should test.
Lastly, please test boundary points which are $(0, 0), (2, 0)$ and $(0, 2)$.
Maximum is $10$ at points $(2, 0)$ and $(0, 2)$ and minima is $1$ at point $(1, 1)$.
A: Write $\mu = \frac{2}{3} \lambda$, the equations becomes
\begin{align} 
\mu x &= x^2 - y , \\
\mu y &= y^2 - x, \\
x^2 + y^2 &=4.
\end{align}
From the first two, we have
$$ \mu (x-y) = (x-y)(x+y) + (x-y),$$
thus
$$ (x-y) ( \mu -1 -x-y) = 0.$$
Then either $x=y$, or $x + y = \mu-1$.
In the case $x=y$, the third equation would give you two points. In the second case, we add the first two equations to get
$$ (\mu +1)( x+y) = 4. $$
Then $\mu$ is found (which has two values) and from there it is not difficult to find $(x, y)$ using
\begin{align}
x+y&= \frac{4}{1+\mu}, \\
x^2 + y^2 &= 4.
\end{align}
