This problem was given in 2018 in the entrance exams for MSU(8th problem). The only information online about the exam is its answers and is supposedly aimed at highschool students. I am in a maths-oriented high school so we are studying what we would in the 1st year in uni. The problem is:
Find all pairs of $x,y \in (0;\frac{\pi}{2}) $ where $f(x,y) = f_{min}$. (When is this equal to its minimum value)
$$ f(x,y) = \left(\frac{\sqrt3\sin y} {\sqrt2\sin(x+y)} + 1\right)\left(\frac{\sqrt2\sin x}{3\sin y} + 1\right)^2\left( \frac{\sin(x+y)}{7\sqrt3\sin x} + 1\right)^4 $$
I only want a starting point, how do I approach this? Derivatives? Or do the (…)^4 ?
Best Answer
I think maybe the question is poorly worded. It says "find all pairs" but provably finding all solutions to the stationary point equations is tedious (even a CAS struggles). Allow me to reword the question:
Find the unique stationary point of the function $f(x, y)$ in the given interval.
What difference does that change in wording make? It means that you don't need to directly find all solutions and can just guess at solutions until you find one that works. Bear that in mind and I will now hint but not explicitly give the solution.
The function you want to find the stationary point of is of the form $f(x, y) = a(x, y)b(x, y)^2c(x, y)^4$ so using $f_x$ for $\frac{\partial f}{\partial x}$ we have $$ f_x = a_x b^2 c^4 + 2 b_x a b c^4 + 4 c_x a b^2 c^3 $$ but the key is we don't have to calculate this whole expression explicitly. Rather we recognise that it is zero if each term has a zero factor. The same goes for $f_y$.
Now the big hint: if you calculate $a_x$ etc what you will see is that the terms in $f_x$ and $f_y$ have some common factors. We just need to find $x$ and $y$ to make those common factors zero and then we have found a stationary point of $f$.
Following the above I find the same solution as in Claude Leibovici's answer.