Find conditional extremum of $f(x, y) = x^4 + y^4$ with condition $(x - 1)^3 - y^2 = 0$
I tried to use Lagrange multipliers: $$L= x^4 + y^4 + \lambda(x^3 - 3x^2 + 3x - 1 - y^2)$$
I derived the system
$\begin{cases} 4x^3 + 3\lambda x^2 - 6\lambda x + 3\lambda = 0\\ 4y^3 - 2\lambda y = 0\\ (x - 1)^3 - y^2 = 0 \end{cases}$
So, one solution is $y = 0$, $x = 1$, but $\lambda$ could by arbitrary. So I don't know how to handle this case.
The other solution is $y^2 = \frac{\lambda}{2}$, and for $x$, I can't express it $x$ in terms of $\lambda$.
Can you help me finish the problem?
There is a cusp at (1,0).
Now, consider the first quadrant, where the gradient vector of $x^4 + y^4$ has both components positive; the $x$ partial is positive, so is the $y$ partial derivative. Such a vector cannot be orthogonal to the curve. Indeed, you wrote it as $(x-1)^3 - y^2,$ the gradient is $\langle 3 (x-1)^2, - 2 y \rangle, $ on the portion of the curve in the first quadrant the $x$ partial is positive but the $y$ partial negative. This cannot be parallel to the other gradient vector, $\langle 4x^3, 4 y^3 \rangle.$ Also, as you can see, $x^4 + y^4$ is actually unbounded on the curve. Just let $x = 1 + t^2, y = t^3$ as $t \rightarrow +\infty$