<s>[INST] Absolute Extrema Multivariable Help Please I been struggling with question for quite a while now and wanted some help. The question is to find the absolute max and min of the function $f(x,y) = x^3+y^3-3x-3y+2$,  bounded by the region $x^2+y^2 \le 1$. I first found the partial derivatives and solved them by setting them zero. However, the values are inconclusive since the coordinates lie outside the region. This is where I get stuck. I tried by substituting $x^2=1-y^2$, but I keep getting the wrong answer. Then I tried using parameterization by setting $x=\cos t$ and $y=\sin t$, but I continue to get the wrong answer. Can you please help me.
Thank you
P.S. Sorry for the equations I am new to the site and this is my first question. 
 [/INST] Let $x = r\cos \theta, y = r\sin \theta$ with $0 \le r \le 1, 0 \le \theta \le 2\pi$. Also, let $u = \cos \theta + \sin \theta, v = \sin \theta \cos \theta$. Then:$ |u| \le \sqrt{2}, x^3 + y^3 = r^3(\cos^3 \theta + \sin^3 \theta)=r^3(\cos \theta+\sin \theta)(1-\sin \theta \cos \theta)= r^3u(1-v)$. And we have: $u^2 - 2v = 1\implies v = \dfrac{u^2-1}{2}\implies x^3+y^3 = \dfrac{r^3u(3-u^2)}{2}\implies f(x,y) = \dfrac{1}{2}r^3u(3-u^2) - 3ru+2= g(r,u) + 2$. We can find the min and max values of $g$, and then add $2$ to get the min and max values of $f$. We use the traditional "critical points" method to find the relative extrema for $g$. We have: $\dfrac{\partial g}{\partial r} = 0 \iff \dfrac{3r^2u(3-u^2)}{2} - 3u = 0$, and $\dfrac{\partial g}{\partial u} = 0 \iff \dfrac{r^3(3-3u^2)}{2}-3r = 0$. Solving this system of equations in $r,u$ we have the solutions: $(u,r) = (0,0), (0,\sqrt{2}), (0,-\sqrt{2})$ of these only the point $(0,0)$ is a valid solution since $0 \le r \le 1$. Thus the only critical point for $g$ is the origin $(0,0)$, and $g(0,0) = 0$. We now check the boundary points that are points on the $4$ sides of the rectangular in the $u$-axis (horizontal), and $r$-axis (vertical) whose vertices are: $A = (-\sqrt{2}, 0), B = (-\sqrt{2}, 1), C = (1,\sqrt{2}), D = (\sqrt{2}, 0)$. On the segment $AB$, we have $g(r,u) = g(r,-\sqrt{2})= 3\sqrt{2}r - \dfrac{r^3\sqrt{2}}{2}$, has $g'(r) = 3\sqrt{2}-\dfrac{3r^2\sqrt{2}}{2} > 0$. Thus on $AB$, $g_{\text{min}} = g(0,-\sqrt{2}) = 0, g_{\text{max}} = g(1,-\sqrt{2}) = \dfrac{5\sqrt{2}}{2}$, you can do the same for $BC$, $CD$, and $DA$, and the min of these values of $g$ together with $0$ is the global min, and the max values of these values together with $0$ is the global max of $g$, then add $2$ to them to get the global min and max for $f$. 
Note: The method that you chose still works since if the critical point does not lie in the domain, then you just consider the boundary points which is points on the unit circle $x^2+y^2 = 1$, and you can use Lagrange Multiplier or the traditional critical points methods that both work !.
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