We look at $f(x, y, z) := xy + z^2$ on unit circle $B := \{(x, y, z) \in \mathbb{R}^3, x^2 + y^2 + z^2 \leq 1\}$.
A) Explain why $f$ has a minimum and maximum on $B$.
B) Show that $f$ has no local extrema on the internal $\{(x, y, z) \in \mathbb{R}^3, x^2 + y^2 + z^2 < 1\}$.
C) Calculate the minimum and maximum of $f$ with the method of Lagrange on $\{(x, y, z) \in \mathbb{R}^3, x^2 + y^2 + z^2 = 1\}$.
My ideas:
A) I don't know how to explain this properly..? I believe the sphere and the function intersect?
B) For $\nabla f (a) = 0$ is $a$ the local extremum.
We have \begin{align} f_x = y \\ f_y = x \\ f_z =2z. \end{align} So $y = 0, x = 0, z = 0$ and to see if the critic point is a maximum we need $f_xx < 0$. $f_xx = 0 \nless 0$, so it can't be a maximum and $ 0 \ngtr 0$ so it can't be a minimum either.
C) For Lagrange: $\nabla f - \lambda \nabla g =0$. Say $g(x, y, z) = x^2 + y^2 + z^2 -1$ is the restriction, then we have \begin{align} f_x = y \qquad g_x = 2x \\ f_y = x \qquad g_y=2y\\ f_z =2z \qquad g_z = 2z. \end{align} This gives: $ y - 2x\lambda = 0, x - 2y\lambda = 0, 2z - 2z \lambda = 0$, which leads to: \begin{align} z = 0 \qquad x^2 = y^2, \text{so} \hspace{2mm} x = y. \end{align} This gives the points $(\sqrt(\frac{1}{2}),\sqrt(\frac{1}{2}), 0)$ and $(-\sqrt(\frac{1}{2}), -\sqrt(\frac{1}{2}), 0)$, which gives the same maximum as minimum??
For point A, notice that $f$ is continuous and $B$ is a closed and bounded set. Therefore $f$ attains a minimum and maximum on $B$.
For C, you are missing some extrema. Notice that when $\lambda=1$, $z$ doesn't necessarily have to equal $0$. When $\lambda=1$, we still have $x=y$, so we can parameterize everything in terms of $x$ to maximize the objective. With this parameterization, our objective becomes $f(x,x,\sqrt{1-2x^2}) = x^2 + (1-2x^2) = 1-x^2$, which attains a maximum value at $x=0$, which implies $y=0$ and $z = \pm1$. So there are two more extrema, $(0,0,1)$ and $(0,0,-1)$. The value of the objective on these extrema is $1$, so these are maxima, whereas the objective takes on the value $1/2$ at the extrema you found, so they are minima.