Langrarian multiplier

Question

Consider the following function
$$f(x, y)=x^4-y^2$$ And Set $A=\{(x,y)\in R^2: x^2+y^2=1\}$ is required.

1. find the Lagrangian equation that determines the extreme point of $F$ on $A$ and calculates the solution for this equation.
2. characterize the above equation into local maxima and minima.
3. using second order condition explain if there are global maxima and minima.

Answer 1

$\nabla (x^4 - y^2) = \nabla(\lambda(x^2+y^2 - 1))$

$4x^3 = 2\lambda x\\ -2y = 2\lambda y$

From the second equation we get: $\lambda = -1 \text { or } y = 0$

if $\lambda = -1\\ 4x^3 = -2x\\ x = 0$

Maxima -- $(1,0),(-1,0)$

Minima -- $(0,1),(0,-1)$

Answer 2

We know that $$\nabla f=(4x^3,-2y)$$and$$c(x,y)=x^2+y^2-1=0$$therefore the Lagrangian would be $$L(x,y)=\nabla f+\lambda\nabla c=(4x^3,-2y)+\lambda(2x,2y)$$also the Hessian of the function is $$H_f=\begin{bmatrix}12x^2&0\\0&-2\end{bmatrix}$$for finding maxima and minima we have $$4x^3=2\lambda x\\-2y=2\lambda y$$if $y=0$ we have $x=\pm 1$ and $\lambda=2$. The only feasible direction in these points is $\vec d=(0,d_2)$ for any $d_2\in\Bbb R-\{0\}$ from the other side$$dH_f(\pm 1,0)d^T=-2d_2^2<0$$which means that the points $(\pm 1,0)$ are local maxima.

If $\lambda=-1$ we obtain $$4x^3=-2x$$ the only possible candidates are $(0,\pm 1)$ that both are local minima similarly. Here is a sketch: