Consider the function $z = 3(y+1)^2-2x^2$ on the domain bounded by the curves : $y = x$ and $ y = x^2-6 $
Sketch the domain.
Find and classify all critical points.
Find the global maximum and minimum.
I don't know how to solve this problem and I don't know any online resources that could help me learn this. Specifically problems bounded by two curves, help is greatly appreciated.
The domain (region between $y=x$ and $y=x^2-6$):
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The critical points of $z = 3(y+1)^2-2x^2$: $$\begin{cases}z_x = -4x=0 \\ z_y=6y+6=0\end{cases} \Rightarrow (x,y)=(0,-1) \quad \text{feasible}.$$ The critical points on the border $y=x$: $$z(x)=3(x+1)^2-2x^2 \Rightarrow z_x=2x+6=0 \Rightarrow \\ (x,y)=(-3,-3) \quad \text{nonfeasible}$$ The critical points on the border $y=x^2-6$: $$z(x)=3(x^2-5)^2-2x^2 \Rightarrow z_x=6(x^2-5)(2x)-4x=0 \Rightarrow \\ (x,y)=(0,-6) \ \text{feasible}; \left(\pm \sqrt{\frac{16}{3}};\frac23\right)\approx (\pm5.33;-0.67) \ \text{nonfeasible}$$ The corner points: $$(x,y)=(-2;-2); (3;3) \ \text{feasible}$$ Can you check all feasible points and determine the maximum and minimum value of the objective function?