This is not supposed to be a 'how-to-do' question, but rather a 'why' question. I came across the following problem:
$Let\ f: R^2 \rightarrow R;\ \ f(x,y)=x^3+y^3.\ Find\ the\ extrema\ points\ of\ f\ so\ that\ x^2+y^2=1$
The solution for the minimum works in the same way as that for finding the maximum, so I'll just consider the maximum when explaining my approach. I took the Lagrange function, made its partial derivatives equal to $0,\ $ found $\ \lambda\ $ so that $g(x,y)=x^2+y^2-1=0\ $ which resulted in some possible solutions. What I find strange is when checking the solution of this problem $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\ $ I found given as a maximum point, whereas $(1,0)$ (and, of course, $(0,1))\ $ clearly yields a greater $\ f$ while also verifying $g(x,y)=0$ and all the partial differential tests for $\ \lambda=-\frac{3}{2}$. Still, $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})\ $ passes the Hessian test while $(1,0)$ doesn't, but this just leaves the Hessian inconclusive for $(1,0)$, which shouldn't be a problem.
So which is the maximum of $\ f$ on $\ x^2+y^2=1?$ $$(1,0)\ and\ (0,1)\ or\ (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})?$$
with $$y=\pm\sqrt{1-x^2}$$ you will get $$f(x,\pm\sqrt{1-x^2})=x^3+(\pm\sqrt{1-x^2})^3$$ and you Problem is reduced to a Problem in only one variable.