Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2006 / Paper 1 / 8

For the Lagrange multiplier equations, use $\mathcal{L}=x^{2}y+2y-\lambda (x^2+y^2-1)$. They give $2x(y-\lambda )=0$, $x^2+2=2\lambda y$, and $x^2+y^2=1$. If $x\ne 0$, then $\lambda =y$, implying $x^2+2=2y^2$ and therefore $y^2=1$, contradicting $x\ne 0$. Thus only $x=0$, $y=\pm 1$ remain.

The circle is compact and the function is continuous, so global extrema exist. On it, eliminating $x^2$ gives $f=3y-y^3$, which is increasing for $-1<y<1$ since its derivative is $3(1-y^2)>0$. ThereforeThese constrained extrema are flat to quadratic order along the circle, so a vanishing second derivative alone would not classify them.

Minimize $\frac{1}{2}\Vert \mathbf{r}{\Vert }^2$ subject to ${\mathbf{n}}_1\cdot \mathbf{r}=1$, ${\mathbf{n}}_2\cdot \mathbf{r}=2$, where ${\mathbf{n}}_1=(2,1,-1)$ and ${\mathbf{n}}_2=(1,-1,1)$. The Lagrange multiplier equation gives $\mathbf{r}=\lambda {\mathbf{n}}_1+\mu {\mathbf{n}}_2$. Since ${\mathbf{n}}_1\cdot {\mathbf{n}}_2=0$, $\Vert {\mathbf{n}}_1{\Vert }^2=6$, and $\Vert {\mathbf{n}}_2{\Vert }^2=3$, the constraints imply $\lambda =1/6$, $\mu =2/3$. HenceNo explicit parametrization of the line was needed. For any other feasible point, $\mathbf{h}=\mathbf{r}-{\mathbf{r}}_{\ast }$ is orthogonal to both normals and therefore to ${\mathbf{r}}_{\ast }$. The Pythagorean theorem gives $\Vert \mathbf{r}{\Vert }^2=\Vert {\mathbf{r}}_{\ast }{\Vert }^2+\Vert \mathbf{h}{\Vert }^2$, proving the unique global minimum. This is the geometric principle behind the minimum-norm least-squares solution.