Past exam of the year 1 of the computer science course of the University of Oxford / 2015 / Discrete mathematics, continuous mathematics, and probability (resit) / 5 / c

Let $g_1=x^2+y^2+z^2-4$ and $g_2=(x-3{)}^2+y^2+z^2-4$. Subtracting the two constraint equations gives $-6x+9=0$, hence $x=3/2$ and $y^2+z^2=7/4$. The feasible set is therefore a circle.

For Lagrange multipliers $\lambda ,\mu$, the stationarity equation $\nabla f=\lambda \nabla g_1+\mu \nabla g_2$ becomesAt $x=3/2$, the first equation gives $\lambda =\mu$. The second ensures $\lambda +\mu \ne 0$ and $y\ne 0$, so the third gives $z=0$. The constraints then force $y=\pm \sqrt{7}/2$. Both points do satisfy the multiplier equations, taking $\lambda =\mu =1/(4y)$.

There are no missing singular constraint points: on the feasible circle, the two constraint gradients $(3,2y,2z)$ and $(-3,2y,2z)$ are linearly independent, since $y^2+z^2=7/4>0$. Thus the complete list of constrained critical points and critical values isThe bound $|y|\le \sqrt{7}/2$ on the feasible circle shows that the positive value is the global maximum and the negative value is the global minimum.