Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 4C (Variational Principles)

For a twice differentiable function, convexity on ${\mathbb{R}}^2$ is equivalent to its Hessian matrix being positive semidefinite everywhere. Here the Hessian matrix is constant:Its diagonal entries are positive, so positive semidefiniteness is equivalent to $\det H=8-{\alpha }^2\ge 0$. The required parameter range isThe endpoints give convex functions with one flat direction; strict inequalities give a Hessian matrix defining a positive-definite quadratic form and a strictly convex function.

Completing the square in the optimization constraint givesThe left side is nonnegative, so every feasible point has $x_1^2\ge 4$. Equality is feasible precisely when $x_2=3x_1/2$ and $x_1=\pm 2$. Thus the global minimum isThere is no maximum: for every $|x_1|\ge 2$, the values $x_2=3x_1/2\pm \sqrt{x_1^2/4-1}$ satisfy the optimization constraint, allowing $x_1^2$ to be arbitrarily large. This direct argument also establishes that the two stationary points are the only local minima on the two branches of the constraint curve.