Analytically minimize $x^2 + y^2$ constrained to $(x+1)^3 =- y^2$ I have the following problem:
$$\text{min} \ x^2 + y^2$$
$$s.t. \ (x+1)^3 =- y^2$$
What I did was substituting, so I got the function $f(x) = x^2  - (x+1)^3 $ but I don't know how to get analytically to the minimum. Graphing the original problem gives $x = -1$ and $y=0$ as the solution but I can't get there using my new function, I did the first derivative equal to $0$ but that has no real solutions. I did Lagrange method on the original problem but it was nonsense too. Thanks for the help.
 A: The constraint $(x+1)^3 = -y^2$ does more than just let you replace the objective function by $x^2 - (x+1)^3$: it also lets you know that $(x+1)^3 \le 0$ (otherwise there is no value of $y$ to correspond to the $x$). Therefore $x+1 \le 0$, or $x \le -1$.
Since $f(x) = x^2 - (x+1)^3$ has derivative $f'(x) = 2x - 3(x+1)^2$, which is always negative, $f$ is always decreasing, so we should set $x=-1$: the highest it can go.
When $x=-1$, $y=0$, and $x^2+y^2=1$.
A: We have that
$$x^2+y^2\ge 0$$
and by the constraint we need $x\le -1$.
Therefore since $$f’(x)=2x-3(x+1)^2<0$$
the minimum is $f(-1)=1$.
A: The constraint curve $ \ (x+1)^3 + y^2 \ = \ 0 \ $ is a "degenerate" elliptic curve (or "semi-cubic" curve) opening away from the $ \ y-$axis with a cusp at $ \ (-1 \ , \ 0 ) \ $ as you found.  Since the function is the "distance-squared from the origin" function, the absolute minimum value will be the distance-squared to that point equal to $ \ 1 \ \ ; $ because the curve extends indefinitely into the second and third quadrants, there is no absolute maximal value.
The Lagrange equations are
$$ 2x \ = \ \lambda · 3·(x+1)^2 \ \ , \ \ 2y \ = \ \lambda · 2y \ \ . $$
The second equation should be factored to obtain $ \ 2y · (1 - \lambda) \ = \ 0 \ \Rightarrow \ y = 0 \ \ \text{or} \ \ \lambda = 1 \ \ . $
The $ \ y = 0 \ $ solution gives you the point you already know about. The difficulty pointed out in some other answers arises from the fact that the cusp does not have a derivative there.  (Edit: 8/19) While we can write an expression for the gradient of the constraint curve as $ \ \nabla g \ = \ \langle \ 3(x+1)^2 \ , \ 2y \ \rangle \ \ , $ we find it to be $ \ \langle 0 \ , \ 0 \rangle \ \ , $ the zero vector, at $ \ (-1 \ , \ 0 ) \ \ . $  It is not possible for the level-curves of the function $ \ x^2 + y^2 \ \ , $ which are circles, to be tangent to that cusp.  A limitation of the Lagrange method is "exposed" in this way.
[ADDENDUM: What we would generally do on finding such a "critical point" is to investigate it individually.  We then discover that it is the closest point to the origin, but we won't learn that from the Lagrange method alone.]
For $ \ \lambda = 1 \ \ , $ inserting this into the second equation produces $ \ 2x - 3·(x+1)^2 \ = \ 0 \ \ , $ which already looks unhelpful.  Indeed, we obtain $ \ 3x^2 + 4x  + 3 \ = \ 0 \ \ $ with the negative discriminant $ \ 16 - 4·3·3 \ \ . $  So there is only the one extremal point at $ \ (-1 \ , \ 0) \ \ . $
A: With Lagrange multipliers:
$$F(x,y) = x^2 + y^2 - \lambda\, (\ (x+1)^3 + y^2 ) $$
The associated system:
\begin{eqnarray}
 \frac{\partial F}{\partial x} &=& 2 x - \lambda\cdot 3 (x+1)^2 = 0 \\
 \frac{\partial F}{\partial y} &=& 2 y - \lambda\cdot  2 y  = 0 \\
 \frac{\partial F}{\partial \lambda}&=&- (\ (x+1)^3 + y^2) =0
\end{eqnarray}
From the second equation: $(\lambda-1)y = 0$. Now, $y$ cannot be $0$, since otherwise from $3$-rd we get $(x+1) = 0$, but then the $1$-st equation is not satisfied.  Therefore, $\lambda = 1$, so
$2 x - 3(x+1)^2=0$. This equation has no real solutions.  So what is the problem?  The set defined by $(x+1)^3 + y^2 = 0$ is closed, so it has at least one closest point on the origin.  The problem is that the function $(x+1)^3 + y^2$ has a (unique) singular point $(-1, 0)$. Therefore, this must be the closest point.
A: Believe in Algebra
If $x+1>0$, then we get a contradiction: $y^2<0$.
This immediately implies, $x+1≤0\implies x≤-1$ must be.
Then, we can define the function $f(x)$, such that  $$f(x)=x^2-(x+1)^3,~x≤-1.$$
Method $-1$
$$\begin{align}x≤-1\implies &\begin{cases}x^2≥1\\ (x+1)^3≤0\end{cases}\\\\
\implies &\begin{cases}x^2≥1\\ -(x+1)^3≥0
\end{cases}\\\\ \implies &x^2-(x+1)^3≥1.\end{align}$$
Finally, we conclude that,
$$\begin{align}&\min \left\{x^2-(x+1)^3\mid x≤-1\right\}=1,\\
&~~~~~~~~~~~~\text{at}~|x|=1,~x+1=0 \\ \\
\implies &\min \left\{x^2-(x+1)^3\mid x≤-1\right\}=1,\\
&~~~~~~~~~~~~~~~~~~~~~\text{at}~x=-1\\\ \end{align}$$
This means,
$$\min \left\{x^2+y^2 \mid (x+1)^3=-y^2\right\}=1,\\ 
\text{at}~x=-1,~y=0.$$

Method $-2$
$$\begin{align}x^2-(x+1)^3&=(x+1)^2-2x-1-(x+1)^3\\&=(x+1)^2-2(x+1)+1-(x+1)^3\\
&=(x+1)^2-(x+1)^3-2(x+1)+1\\
&=z^3+z^2+2z+1≥1,~z=-(x+1)≥0.\\ \end{align}$$

Method $-3\text{a}$
$$\begin{align}x^2-(x+1)^3&=x^2-1-(x+1)^3+1=\\
&=(x-1)(x+1)-(x+1)^3+1\\
&=(x+1)(x-1-x^2-2x-1)+1\\
&=\underbrace{-(x+1)}_{≥0}\underbrace{(x^2+x+2)}_{>0}+1≥1.\end{align}$$
Method $-3\text{b}$
$$\begin{align}x^2-(x+1)^3&=-(x+1)(x^2+x+2)+1\\
&=-(x+1)\left((x+1)^2-(x+1)+2)\right)+1\\
&=-(x+1)^3+(x+1)^2-2(x+1)+1\\
&=z^3+z^2+2z+1≥1,~z=-(x+1)≥0.\end{align}$$
A: A Pre-Calculus Approach
Since $(x+1)^3=-y^2$, we know that $x\le-1$. Furthermore,
$$
\begin{align}
x^2+y^2
&=\overbrace{1-2(x+1)+(x+1)^2}^{x^2=(1-(x+1))^2}-(x+1)^3\tag1\\
&=1+2u+u^2+u^3\tag2\\
&\ge1\tag3
\end{align}
$$
Explanation:
$(1)$: $y^2=-(x+1)^3$
$(2)$: substitute $u=-(x+1)$
$(3)$: $u\ge0$ since $x\le-1$
Since $(x,y)=(-1,0)$ satisfies $(x+1)^3=-y^2$ and gives $x^2+y^2=1$, we see that $1$ is the minimum of $x^2+y^2$.


A: Consider the function $f(x) = x^2 - (x+1)^3\implies f'(x)=2x-3(x+1)^2= -(3x^2+4x+3) < 0$ for all reals $x$. Thus letting $x \to +\infty$, $f(x) \to -\infty$. This means the minimum value does not exist, and neither is the maximum.
A: EDIT: (graph corrected)
Euler-Lagrange leads to
$$3x^2+4x+3=0 \text{
with complex roots }  ((-2+\sqrt i)/3, z_2)$$
There is no regular point intersection. Graphically seen the biggest radius of circle to contact the constraining condition has unit radius at its cusp point $(-1,0)$.
Note that the contact maximizes but does not minimize all possible  circle radii from origin.

