There is two light bulbs, one on $(-a,0,2)$ and the other on $(a,0,2)$, where $a>0$.
The light intensity coming from two light bulbs on the $x$-axis and $y$-axis (where $z=0$) can be calculated using the following equation: $$f(x,y) = \frac{1}{(x+a)^2+y^2+2^2} + \frac{1}{(x-a)^2+y^2+2^2}.$$ My question is, how to find the maximum point(s) (or critical points) of $f$.
Notes: There can be one or two critical points, depending on $a$.
$\def\peq{\mathrel{\phantom{=}}{}}$For any critical point $(x_0, y_0) \in \mathbb{R}^2$,$$ \frac{\partial f}{\partial x}(x_0, y_0) = \frac{\partial f}{\partial y}(x_0, y_0) = 0. $$ Because for any $(x, y) \in \mathbb{R}^2$,\begin{align*} \frac{\partial f}{\partial x}(x, y) &= -2 \left( \frac{x + a}{((x + a)^2 + y^2 + 2^2)^2} + \frac{x - a}{((x - a)^2 + y^2 + 2^2)^2} \right),\\ \frac{\partial f}{\partial y}(x, y) &= -2 \left( \frac{y}{((x + a)^2 + y^2 + 2^2)^2} + \frac{y}{((x - a)^2 + y^2 + 2^2)^2} \right), \end{align*} then $\dfrac{\partial f}{\partial y}(x_0, y_0) = 0 \Rightarrow y_0 = 0$, and\begin{align*} &\mathrel{\phantom{\Longrightarrow}}{} \frac{\partial f}{\partial x}(x_0, y_0) = \frac{\partial f}{\partial x}(x_0, 0) = 0\\ &\Longrightarrow \frac{x_0 + a}{((x_0 + a)^2 + 2^2)^2} + \frac{x_0 - a}{((x_0 - a)^2 + 2^2)^2} = 0\\ &\Longrightarrow (x_0 + a)((x_0 - a)^2 + 2^2)^2 + (x_0 - a)((x_0 + a)^2 + 2^2)^2 = 0. \tag{1} \end{align*} Also, for any $x_0 \in \mathbb{R}$ satisfying (1), $(x_0, 0)$ is indeed a critical point. Since for any $x \in \mathbb{R}$,\begin{align*} (x + a)((x - a)^2 + 2^2)^2 &= (x + a)(x - a)^4 + 8(x + a)(x - a)^2 + 16(x + a)\\ &= (x^2 - a^2)(x - a)^3 + 8(x^2 - a^2)(x - a) + 16(x + a), \end{align*} then\begin{align*} &\peq (x + a)((x - a)^2 + 2^2)^2 + (x - a)((x + a)^2 + 2^2)^2\\ &= (x^2 - a^2)(x - a)^3 + 8(x^2 - a^2)(x - a) + 16(x + a)\\ &\peq + (x^2 - a^2)(x + a)^3 + 8(x^2 - a^2)(x + a) + 16(x - a)\\ &= (x^2 - a^2) \Bigl((x - a)^3 + (x + a)^3 + 8(x - a) + 8(x + a)\Bigr) + 32x\\ &= (x^2 - a^2)(2x^3 + 6a^2 x + 16x) + 32x = 2x \Bigl( (x^2 - a^2)(x^2 + 3a^2 + 8) + 16 \Bigr)\\ &= 2x \Bigl(x^4 + (2a^2 + 8)x^2 - (3a^4 + 8a^2 - 16) \Bigr) = 2x \Bigl( (x^2 + a^2 + 4)^2 - 4a^2 (a^2 + 4) \Bigr). \end{align*} Note that$$ (a^2 + 4)^2 - 4a^2 (a^2 + 4) = -(3a^4 + 8a^2 - 16) = -(a^2 + 4)(3a^2 - 4). $$ If $(a^2 + 4)^2 - 4a^2 (a^2 + 4) \geqslant 0$, i.e. $0 < a \leqslant \dfrac{2}{\sqrt{3}}$, then $(1) \Leftrightarrow x_0 = 0$. If $(a^2 + 4)^2 - 4a^2 (a^2 + 4) < 0$, i.e. $a > \dfrac{2}{\sqrt{3}}$, then$$ (1) \Longleftrightarrow x_0 = 0 \text{ or } \pm \sqrt{a^2 + 4} \sqrt{2a - \sqrt{a^2 + 4}}. $$ Therefore, if $0 < a \leqslant \dfrac{2}{\sqrt{3}}$, then the only critical point is $(0, 0)$. If $a > \dfrac{2}{\sqrt{3}}$, then the critical points are $(0, 0)$ and $(\pm \sqrt{a^2 + 4} \sqrt{2a - \sqrt{a^2 + 4}}, 0)$.