I have the following problem:
Let $B$ be the set $B=\{(x_1 , x_2) \in \mathbb{R}^2\mid x_1^2 + x_2^2 \leq 1\text{ and }x_1 \geq 0\}$.
A function $f:B \rightarrow \mathbb{R}$ is given by $$ f(x_1 ,x_2 )=x_1^2 x_2^2 +3x_1 x_2 +x_2 -4.$$ State the image of $f$.
To find the image, I want to do an extrema investigation.
First, I find all the stationary points. There is one at $\left(-\frac13, 0\right)$, but based on the Hessian matrix, I conclude it is a saddle point.
Next, I want to do a boundary investigation. I parametrize the circular part of the boundary of the set as
$$s(u) = (\cos u,\sin u),\quad u \in\left[-\frac\pi2,\frac\pi2\right].$$
However, this is where my problem starts. I make a composite function $f \circ s$ - this describes the values of $f$ along the circular boundary. I then differentiate this composite function and get
$$- 2 \sin^3u\,\cos u- 3 \sin^2u+ 2 \sin u\,\cos^3u+ 3 \cos^2u+ \cos u.$$
Now, I want to set it equal to $0$ and solve, since the stationary points on the composite function constitute possible extrema of $f$ on $B$. However, I am not able to solve this. It there any trick or another way to find these extrema?
Considering turning points, there is one at $(-1/3,0)$. However we must also check on the boundary of the set for further extrema. If you take $x_2 = \pm \sqrt{1-x_1^2}$ and then look for turning points along this boundary, using numerics, you should find the minimum to be $-6.0069$, and the maximum to be $-1.51414$. It would be nice to do it without numerics but the derivative is tricky. Perhaps someone else can shed light on how to solve the derivative $= 0$.
If $x_2 = \pm \sqrt{1-x_1^2}$ then set $g(x) = f(x_1,x_2) = x^2(1-x^2)\pm (3x+1)\sqrt{1-x^2}-4$.
$g'(x) = 2x-1/3x^3\pm3\sqrt{1-x^2}\pm(3x+1)\frac{1}{\sqrt{1-x^2}}$