Determine if the function $f (x, y) = x^4 − y^2 − x^2 + y$ has a maximum or minimum value for $(x, y)$ in the range $D = \{(x, y) : x^2 \leq y \leq 1\}$ and, where appropriate, determine these.
This is how I tried solving the question: One thing I realised is that our function is bounded by $g_1(x,y) = x^2-y$ and $g_2(x,y) = y-1$. Lets focus on $g_2$ right now, at least that is what I did. Here:
I found the Lagrange function:
$L(x,y,z) = x^4 − y^2 − x^2 + y - λ(x^2 - y)$
$\nabla L = (4x^3 - 2x -2\lambda x, -2y+\lambda + 1, -x^2+y) = (0,0,0)$
and I get the following system of equations:
$${ \begin{cases}{4x^3 - 2x -2\lambda x = 0} \\ {-2y+\lambda + 1 = 0} \\ {-x^2+y = 0} \end{cases} }$$
$$x(4x^2 - 2 - 2\lambda)= 0 \iff x = \pm \sqrt{\frac{1+\lambda}{2}}$$
$$-2y+\lambda + 1 = 0 \iff y = \frac{1+ \lambda}{2}$$
$$ -x^2+y = 0 \iff -(\pm \sqrt{\frac{1+\lambda}{2}})^2 + \frac{1+ \lambda}{2} = -\frac{1+ \lambda}{2}+\frac{1+ \lambda}{2} = 0 $$
As you can see I didn't get any solution for $\lambda$. How do solve this? Btw the critical points are: $(\pm \frac1{\sqrt{2}}, \frac12)$
Note that along the curve $y=x^2$ the function is constant: $$f(x,x^2)=x^4 − (x^2)^2 − x^2 + x^2=0.$$ Indeed solving your system we find $$ { \begin{cases}{2x(2x^2 - 1 -\lambda) = 0} \\ {\lambda= 2y-1} \\ {y=x^2} \end{cases} } \Leftrightarrow { \begin{cases}{4x(x^2 -y) = 0} \\ {\lambda= 2y-1} \\ {y=x^2} \end{cases} } \Leftrightarrow { \begin{cases}{0 = 0} \\ {\lambda= 2y-1} \\ {y=x^2} \end{cases} }$$ So they are all critical points!
Since $D$ is a compact set extreme values exist and order to finish the job you should,
find the critical points inside $D$ by solving $$\nabla f(x,y)=(4x^3-2x,-2y+1)=(0,0)$$
investigate $f$ along the rest of the boundary of $D$, i.e. the segment $\{(x,1):x\in [-1,1]\}$: $$f(x,1)=x^4 − 1 − x^2 + 1=x^2(x^2-1).$$