= Solution
Expand $u=3x^2-2xy+3y^2$. The <Lagrange multiplier> equations for $u-\lambda v$ are $(6-2\lambda)x-2y=0$ and $-2x+6y+\lambda=0$, together with the <constraint> $y=x^2-1/4$. The first two imply $y=(3-\lambda)x$ and $(6\lambda-16)x=\lambda$. The value $\lambda=8/3$ would make the latter equation inconsistent, so division is allowed:
$$
\boxed{x=\frac{\lambda}{6\lambda-16},\qquad y=\frac{\lambda(3-\lambda)}{6\lambda-16}.}
$$
Insert these into $x^2-y-1/4=0$ and multiply by $(6\lambda-16)^2$. Simplification gives $2(3\lambda^3-21\lambda^2+48\lambda-32)=0$, hence
$$
\boxed{3\lambda^3-21\lambda^2+48\lambda-32=0.}
$$
The <gradient> of the <constraint> $(2x,-1)$ is never zero, so the <Lagrange multiplier> condition is necessary at every constrained <extremum>. To establish a global <minimum>, parametrize the whole <parabola> by $y=x^2-1/4$. The restricted <function> becomes
$$
q(x)=3x^4-2x^3+\frac32x^2+\frac12x+\frac3{16}.
$$
It is <coercive>: $q(x)\to\infty$ as $|x|\to\infty$, so it attains a global <minimum>. Moreover $q''(x)=36(x-1/6)^2+2>0$, making it <strictly convex> and ensuring that this <minimum> is unique. This proves more than the permitted assumption of one real multiplier root.
The <level sets> of $u$ are nested <ellipses> centered at the origin. In the rotated coordinates $p=(x+y)/\sqrt2$, $q=(x-y)/\sqrt2$, they have equation $2p^2+4q^2=\text{constant}$, with the longer axis along $x=y$. The smallest one meeting the <parabola> is tangent to it at the unique minimizing point, approximately $(x,y)=(-0.12654,-0.23399)$, where $u\simeq0.15307$ and $\lambda\simeq1.15085$.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2003/paper-2-constrained-ellipse-parabola.png]
{title=The smallest quadratic contour touching a parabolic constraint}
{description=Original level-set plot. The highlighted ellipse is tangent to the parabola at the unique global minimum; larger quadratic contours intersect the constraint.}
Back to article page
