Consider points $A\left(\dfrac{17}{4}; 0; 0\right), B(5; 4; 0)$ and spheres $(C_1)\colon \, x^2 + y^2 + z^2 = 1, (C_2)\colon \, x^2 + (y - 4)^2 + z^2 = 4$. Given points $M$ and $N$ mobile on spheres $(C_1)$ and $(C_2)$ respectively. The minimum value of expression $P = MA + 2NB + 4MN$ lies in the range $$\begin{aligned} A&. [\sqrt{265}; 2\sqrt{265}) &B&. [2\sqrt{265}; 3\sqrt{265})\\ C&. [3\sqrt{265}; 4\sqrt{265}) &D&. [4\sqrt{265}; 5\sqrt{265}) \end{aligned}$$
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
This is the first problem where I actually put the options at the start since it doesn't ask for the exact value, only the range where it snuggly comforts itself in.
Let's try to solve this problem in a two-dimensional space first.
Let $(\cos\alpha; \sin\alpha)$ and $(2\cos\beta; 2\sin\beta + 4)$ be the coordinates of points $M$ and $N$ respectively. Therefore, $$\left\{ \begin{aligned} MA &= \sqrt{\left(\cos\alpha - \dfrac{17}{4}\right)^2 + \sin^2\alpha} = \sqrt{\dfrac{305}{16} - \dfrac{17}{2}\cos\alpha}\\ NB &= 2 \times \sqrt{\left(\cos\beta - \dfrac{5}{2}\right)^2 + \sin^2\beta} = 2 \times \sqrt{\dfrac{29}{4} - 5\cos\beta}\\ MN &= \sqrt{(\cos\alpha - 2\cos\beta)^2 + (\sin\alpha - 2\sin\beta - 4)^2}\\ &= \sqrt{[5 - 4\cos(\alpha - \beta)] - 8(\sin\alpha - 2\sin\beta - 2)}\\ &\ge \sqrt{1 - 8(\sin\alpha - 2\sin\beta - 2)} \end{aligned} \right.$$
We're already assumed that $\cos(\alpha - \beta) = 1 \iff \alpha = \beta + 2n\pi, n \in \mathbb Z \iff \left\{ \begin{aligned} \cos\alpha &= \cos\beta\\ \sin\alpha &= \sin\beta \end{aligned} \right.$, let's just roll on with it.
Let $\cos\alpha = \cos\beta = t \ (-1 \le t \le 1)$, then $\sin\alpha = \sin\beta = \pm\sqrt{1 - t^2}$ and $$P = MA + 2NB + 4MN\\ = \sqrt{\dfrac{305}{16} - \dfrac{17}{2}t} + 4 \times \sqrt{\dfrac{29}{4} - 5t} + 4 \times \sqrt{1 - 8 \times \left(\mp\sqrt{1 - t^2} - 2\right)}$$
For the case of $-\sqrt{1 - t^2} - 2$, the expression $P$ reaches it minimum value at $t = 1$, where it is equal to $\dfrac{37}{4} + 4\sqrt{17} \approx 1.5813444438744137\sqrt{265}$. Why do I know that? Well, it just so happens that my calculator has a function called TABLE where you can input a function and a list of values in an arithmetic progression, you can also control the common difference.
For the case of $\sqrt{1 - t^2} - 2$, this is where it gets tricky, WolframAlpha gives the result of approximately $$24.670565345230251 \approx 1.5155007821107844\sqrt{265}$$ where $t \approx 0.83974334877438$.
In the end, were this problem to be contained in a two-dimensional space, the answer would be $A. [\sqrt{265}; 2\sqrt{265})$. I'd assume that the answer would probably be the same in a three-dimensional space.
Let $(\cos\alpha\sqrt{1 - a^2}; \sin\alpha\sqrt{1 - a^2}; a)$ and $(\cos\beta\sqrt{1 - b^2}; (2\sin\beta + 4)\sqrt{1 - b^2}; b)$ be the coordinates of points $M$ and $N$ respectively $(-1 \le a \le 1, -1 \le b \le 1)$...
But that seems too complicated.
Perhaps you weren't meant to actually figure out the exact value of expression $P$, only an approximation of it. Also, it seems like you need to have an exceptional calculator.
That's all for now. But before we end, I'm wondering. What would the correct answer be were this problem to ask about the maximum value of expression $P$?
Thanks for reading, (and even more so if you could help), and have a wonderful tomorrow~