Let $D \subseteq \mathbb{R}^2$ be defined as follows $$ D = \bigr\{(x,y) \in \mathbb{R}^2 \mid 0 \leq y \leq\sqrt{1-x^2}\bigl\} $$ The problem is to find the maximum and minimum of the function $$f(x,y) = \sqrt{3}x + y$$ for $x,y \in D$.
I tried to let $k = \sqrt{3}x + y$ so that $y = k - \sqrt{3}x$ and then to solve
$$k - \sqrt{3} x = \sqrt{1-x^2}$$
However, this lead to some quite messy calculations. Is there an easier way?
On
The objective function is linear, so the extrema occur at the corner points of the domain. It follows that the maximum value is $\sqrt 3,$ and the minimum $-\sqrt 3.$
On
From what you have,
$$k - \sqrt{3} x = \sqrt{1-x^2}$$
arrange the equation in the standard form
$$4x^2-2\sqrt 3 kx + k^2-1=0$$
Note that $k$ given by the line equation $k=\sqrt 3 x+y$ is the $y$-intercept of the line. $k$ takes the maximum value when it is tangential to the circle from above, that is, the discriminate of the above equation is zero,
$$b^2-4ac = 12k^2-16(k^2-1)=0$$
Thus, the maximum value of $k$ is
$$k_{max} = 2$$
On the other hand, the $y$-intercept $k$ takes the minimum value if the line touches the far-left point $(-1,0)$ on the circle from below, i.e.
$$k_{min}=\sqrt 3(-1)+0=-\sqrt 3$$
How about using angles, $(\cos\theta, \sin\theta) = (x,y)$ ? $$g(\theta) = \sqrt3\cos\theta + \sin\theta$$
$$g'(\theta) = -\sqrt3\sin\theta + \cos\theta$$
For maximum, solve for $g'(\theta)=0$: $$\sqrt3\sin\theta = \cos\theta$$ $$\tan\theta = {1 \over \sqrt3} = \tan({\pi \over 6})$$
$$max(f(x,y)) = \large g({\pi \over 6}) = \sqrt3\times({\sqrt3 \over 2}) + {1\over2} = 2$$ $$min(f(x,y)) = f(-1,0) = -\sqrt3$$
Another way, without doing differentiation
$$g(\theta)=\sqrt3\cos\theta + \sin\theta = 2 \cos(\theta-{\pi\over6})$$
$$max(f(x,y)) = max(g(\theta)) = 2$$