a) Compute the steepest decent path in $\mathbf x_0=(0,0).$
b) Other search paths form an angle with the steepest descent path. For what angle $\alpha$ is the search path a decent direction?
c) Assume that you are solving the patch searching problem exactly using the stepest descent search path. what is the angle $\beta$ between two consecutive search paths?
My attempt:
a) The gradient at $\mathbf x_0$ is computed to $\nabla f=(2x+2y-2,2x+4y)\rvert_{(x,y)=(0,0)}=(-2,0).$ The iterationformula is $\mathbf s_k=-\nabla f(\mathbf x_k)$, so $\mathbf s_0=-(-2,0)=(2,0).$
b) What angle? This $\alpha$ can be measured w.r.t $x,y$ and $z$ direction.
c) I don't understand this question at all.
The gradient $f'=\nabla f$ vanishes at exactly one point, the one with $2x+2y-2=0$ and $2x+4y=0$, which is $P(2,-1)$. So this may be or not a relative extremal point for $f$, but it is the only one, that may be.
Since $f$ has an absolute minimum, this point $P$ must be the one. If this is not "convincing" , then...
Alternatively we can compute the second derivative of $f$ (the Hessian matrix) at that point, which is
$$ f''= \begin{bmatrix} 2 & 2\\ 2 & 4 \end{bmatrix}\ , $$ which is positive definite. So we have a relative minimal point.
To have a quick proof, not a conviction, at low level, now that we know the "to be absolute minimal value", $f(2,-1)=-2$, (if we have this information, we may forget all the above,) it is enough to group squares, and mention the low level estimation: $$ \begin{aligned} f(x,y)-f(2,-1) &= x^2 +2x+2xy+2y^2+2 \\ &= \frac 12 x^2 + 2x+2 +2+\frac 12x^2 +2xy+2y^2 \\ &= \frac 12 (x+2)^2+\frac 12(x+2y)^2 \\ &\ge 0+0=0\ , \end{aligned} $$ with equality, iff $x+2=x+2y=0$.
EDIT: I realized only after the comment below, that we have to use steepest descent. Simplest, and most canonically it is to use the continuous steepest descent, i.e. to search for the (one and only) continuous path $t\to p(t)=(x(t),y(t))$, such that $$ \frac{dp(t)}{dt} = -\nabla f(p(t))\ ,\qquad p(0)=(0,0)\ , $$ i.e. $$ \begin{bmatrix} x'(t)\\y'(t) \end{bmatrix} = - \begin{bmatrix} 2x(t)-2+2y(t)\\ 2x(t)+4y(t) \end{bmatrix} \ . $$ The solution is (by uniqueness, since it satisfies, computer aided,) $$ \begin{aligned} x(t) &= \boxed{+2} -\frac{2}{5} \, \Big(\ 2 \sqrt{5} \sinh(t\sqrt{5}) + 5 \cosh(t\sqrt{5})\ \Big) \exp(-3t) \ , \\ y(t) &= \boxed{-1} +\frac{1}{5} \Big(\ 3\sqrt{5} \sinh(t\sqrt{5}) + 5 \cosh(t\sqrt{5} )\ \Big)\exp(-3t)\ . \end{aligned} $$ (The characteristic polynomial of the associated homogenous polynomial is $\lambda^2+6\lambda +4$, with the roots $-3\pm \sqrt 5$. So we obtain a linear combination of $\exp(t(-3\pm \sqrt 5))$ for it.)
For $t\to\infty$ the "path" converges to $p(\infty)=(-2,1)$, which is the absolute minimum.
In a plot of the above function $p$, and the contours of $f$:
Sage code for it:
As in the post, if we use some discrete variation of the above, than starting from $(0,0)$ we should go in direction $$s_0=(2,0)\ ,$$ which is also the tangent to the red path in the figure. (How far? Depends on book.) From the new point, we have a new continuous path. And so on. It may be that this is related... "Numeric texts" tend generally to be cryptic for me.)
Now in the posted situation, we have some discrete version of the above. The recursion for $p_{k+1}$, starting from $p_k$, should be of the form $$p_{k+1}=p_k\underbrace{-\nabla f(x_k)}_{s_k}\epsilon_k\ . $$ However, there is no relation of the above to some standard notation $\alpha,\beta$ (that i may/should know), so $\alpha,\beta$ are undefined for me, i cannot figure out more, have to stop here. (I think, we need a printed reference, the text may reveal some involuntary hints.)