Finding extrema points with lagrange multipliers Using lagrange multipliers, find all the extrema points of the function $f(x,y) = x^2 + (y-b)^2$ subject to the constraint $y = x^2$. 
Using the fact that critical points occur at $\triangledown f(x,y) = (0,0)$ and so $(2x, 2y-2b) = (0,0)$. So an extrema at $(0,b)$. Should the point $(0,b)$ be included as an extrema since the question asks to use lagrange multipliers to find them? Or is this point outside the constraint $y=x^2$. I know that there are more points that can be found using lagrange multipliers.
Also my lecturer said by giving the parametrization of the parabola $y=x^2$ and substituting this into $f(x,y)$, we can find whether each point found is a local max or a local min. How is this done? 
 A: The critical point (0,b) should not be included as an extrema, since the constraint is $y=x^2$, which is the boundary of the region, whereas (0,b) in inside the region $y=x^2$, so it should not be included.
The way we use to determine local max or local min is through Second Derivative Test.
Suppose the second derivative of f(x,y) are continuous on a open disk, and (a,b) is a critical point, we define $D(a,b)=f_{xx}(a,b)*f_{yy}(a,b)-f_{xy}(a,b)^{2}$, where $f_{xx}$ is the second derivative of f w.r.t x. Then,


*

*If $D(a,b)>0$ and $f_{xx}(a,b)>0$ (or $f_{yy}(a,b)>0)$, then f(a,b) is a local minimum

*If  $D(a,b)>0$ and $f_{xx}(a,b)<0$ (or $f_{yy}(a,b)<0)$, then f(a,b) is a local maximum

*If  $D(a,b)<0$, then f(a,b) is a saddle point

*If $D(a,b)=0$, then the test is inconclusive

A: The geometrical picture is the following: We are asked to find the local extrema of the distance from the point $(0,b)$ on the $y$-axis to points on the parabola $y=x^2$. From looking at a figure we can guess the following: If $b\gg1$ there are two local minima high up, and a local maximum at $(0,0)$. If $0<b\ll1$ there is just one local minimum at $(0,0)$, and the same holds when $b\leq0$.
The intended computation goes as follows: Set up the Lagrangian
$$\Phi:=x^2+(y-b)^2+\lambda(y-x^2)\ ,$$
and solve the system
$$\Phi_x=2x-2\lambda x=0,\quad \Phi_y=2(y-b)+\lambda=0,\quad y=x^2\ .$$
From $x(1-\lambda)=0$ we infer (i) $x=0$ or (ii) $\lambda=1$. In case (i) we then obtain $y=0$ and a certain value of $\lambda$, and in case (ii) we obtain $y=b-{1\over2}$. The condition $y=x^2$ then implies that case (ii)  only leads to real solutions if $b\geq{1\over2}$, and in this case we have $x=\pm\sqrt{b-{1\over2}}$.
It follows that Lagrange's method has confirmed our geometric analysis of the problem. Note however that it is quite cumbersome to do a second derivative test in the framework of this method. Instead we can do the following: Consider the parametric representation $x\mapsto (x,x^2)$ of the parabola, and instead of $f$ plus constraint look at the pullback
$$\psi(x):=f(x,x^2)=x^2+(x^2-b)^2\qquad(-\infty< x<\infty)\ .$$
Now analyze this function $\psi$ as a function of one variable. You will get the same results (depending on $b$) as before, and in addition the second derivative test will confirm what you knew all along. The case $b={1\over2}$ is special: Here the first nonvanishing derivative is $\psi^{(4)}(0)=24$. Since $4$ is even and $24>0$ we have a local minimum there.
