Find the maximum value of $f(x,y)=x^2+y^2$ in the region bounded by $y=\frac x2,y=-\frac x2$ and $x=y^2+1$,including the boundary lines

Question

The figure of the problem where $y=\frac x2,y=-\frac x2$ and $x=y^2+1$ enclose the region is this:

The shaded region is the enclosed area. Now the max of $x^2+y^2$ in this area is exactly what in this region?Is it the point where $y=\frac{x}{2}$ intersects $x^2+y^2$?

Answer 1

If the feasible region is the light blue region then the maximum for $x^2+y^2$ is $\infty$ otherwise if the feasible region is the white one the the maximum is attained at $(x_0,y_0)=(0,1)$ with value $x_0^2+y_0^2 = 1$

Answer 2

The feasible region is the shaded region. The optimal point is at $(1,0)$.

$\hspace{5cm}$

Hence: $$f(1,0)=1 \ (max).$$

Answer 3

$x^2+y^2$ is is square of distance from origin thus the point furtherest from the origin i.e $(2,1)$ and $(2,-1)$ lying on intersection of the parabola and the line are the optimal points with $f_{max}=5$