= Solution
{c}
For $g=(x-1)^2+y^2-29$, the <gradient> $\nabla g=2(x-1,y)$ never vanishes on the <circle>. The <Lagrange multiplier> equations are
$$
2x+y^2+1=2\lambda(x-1),\qquad 2y(1+x)=2\lambda y,\qquad (x-1)^2+y^2=29.
$$
If $y=0$, the two possibilities are $(1\pm\sqrt{29},0)$. If $y\ne0$, then $\lambda=1+x$, and the first equation gives $y^2=2x^2-2x-3$. Combining this with $y^2=28+2x-x^2$ yields
$$
3x^2-4x-31=0.
$$
Put
$$
s=\sqrt{97},\qquad r_-=\frac{2-s}{3},\qquad r_+=\frac{2+s}{3},\qquad h_-=\frac{\sqrt{163-2s}}3,\qquad h_+=\frac{\sqrt{163+2s}}3.
$$
Both $h_-$ and $h_+$ are positive, giving the other four points $(r_-,\pm h_-)$ and $(r_+,\pm h_+)$. To classify all six, restrict $f$ to the <circle> by eliminating $y^2$:
$$
f=H(x)=-x^3+2x^2+31x+28,\qquad 1-\sqrt{29}\leq x\leq1+\sqrt{29}.
$$
Its <derivatives> are $H'=-3x^2+4x+31$ and $H''=4-6x$. In the interior, $H''(r_-)=2s>0$ and $H''(r_+)=-2s<0$. At either endpoint $H'<0$; along the <circle>, $x$ has a strict <local minimum> at the left endpoint and a strict <local maximum> at the right endpoint. Consequently the six constrained <extrema> are
$$
\boxed{\begin{array}{c|c|c}
\text{point}&f&\text{classification on the circle}\\ \hline
(1-\sqrt{29},0)&31-3\sqrt{29}&\text{strict local maximum}\\
(1+\sqrt{29},0)&31+3\sqrt{29}&\text{strict local minimum}\\
(r_-,\pm h_-)&\dfrac{1330-194\sqrt{97}}{27}&\text{global minima}\\
(r_+,\pm h_+)&\dfrac{1330+194\sqrt{97}}{27}&\text{global maxima}
\end{array}}
$$
The <circle> is a <compact set>, so the <extreme value theorem> guarantees that its <global minimum> and <global maximum> occur among these six points. Their values give the corrected sharp bound
$$
\boxed{f(x,y)\geq\frac{1330-194\sqrt{97}}{27}\approx-21.50660791\quad\text{on the circle}.}
$$
\b[The printed lower bound $-20$ is false.] A particularly short counterexample is
$$
(x,y)=\left(-\frac52,\frac{\sqrt{67}}2\right),\qquad (x-1)^2+y^2=\frac{49+67}{4}=29,\qquad f(x,y)=-\frac{171}{8}<-20.
$$
The sole <local minimum> in part (a), $(-1/2,0)$, has value $-1/4$, larger than values on this <circle>; it is therefore \b[not a <global minimum>]. In fact $f(-2,y)=2-y^2\to-\infty$, so $f$ has \b[no <global minimum> on the plane].
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-iii-constrained-extrema.png]
{title=All six extrema on the constraint circle and the corrected minimum of the restricted polynomial}
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