 Solution

All three active constraints force $x_1=2$ and $x_2=4$, which also satisfy $x_1^2+x_2^2=20$. For maximization use $\mathcal{L}=F+{\lambda }^{T}g$, $\lambda \ge 0$. At $x^{\ast }=(2,4)$,Stationarity gives the complete familyIn particular $({\lambda }_1,{\lambda }_2,{\lambda }_3)=(2,3/2,2)$ is strictly positive. Three gradients in two dimensions cannot be linearly independent, so the linear independence constraint qualification fails. But $d=(-1,-1)$ gives $\nabla g_i\cdot d=(1,12,3)>0$, so the Mangasarian–Fromovitz constraint qualification holds.

For a nonzero linearized feasible direction $d$, each $\nabla g_i\cdot d\ge 0$, andThe inequality is strict because equality would require all three inner products to vanish, and the first and third gradients already span ${\mathbb{R}}^2$. This proves a strict local maximum: otherwise a sequence of nearby feasible points with no smaller objective would, after normalizing their differences from $x^{\ast }$ and taking a convergent subsequence, give a unit linearized feasible direction with nonnegative objective derivative, contradicting the displayed inequality. Thus