= 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 $\mathcal L=F+\lambda^Tg$, $\lambda\geq0$. At $x^*=(2,4)$,
$$
\nabla F=(16,10),\quad \nabla g_1=(-1,0),\quad
\nabla g_2=(-4,-8),\quad\nabla g_3=(-4,1).
$$
Stationarity gives the complete family
$$
\boxed{\lambda_2=t,\quad\lambda_3=8t-10,\quad\lambda_1=56-36t,
\qquad\frac54\leq t\leq\frac{14}{9}.}
$$
In 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 $\nabla g_i\cdot d\geq0$, and
$$
\nabla F\cdot d=-2\nabla g_1\cdot d-\tfrac32\nabla g_2\cdot d-2\nabla g_3\cdot d<0.
$$
The 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 $x^*$ and taking a convergent subsequence, give a unit linearized feasible direction with nonnegative objective derivative, contradicting the displayed inequality. Thus
$$
\boxed{x^*=(2,4)\text{ is a strict local maximizer},\quad F(x^*)=32.}
$$
Back to article page
