Determine the greatest possible value of $\sum_{i=1}^{10} \cos(3x_i)$ for real $x_1,x_2,\cdots,x_{10}$, where $\sum_{i=1}^{10} \cos(x_i)=0$.

My approach was to somehow get a equation solely in terms of one of the $\cos x_i$ and then differentiating and getting the maximum according to the range of cos which is $[-1,1]$. For achieving this, I first inductively proved that $$a^3+b^3 +c^3 +\cdots+n^3= 3\cdot(\text{sum of product of terms taking three at a time})$$ but that just makes the expression to be that we need to maximize the value of $$12\cdot(\text{sum of product of terms taking three at a time})$$ But now how do we get an equation in just one variable?


1 Answer 1


We apply Lagrange multipliers.

Let $f(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos 3x_i$ be the objective function and $g(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos x_i=0$ be the constraint.

For each $1\leq i\leq 10$, we have by the triple angle formula,

$\lambda \sin x_i = 3(3\sin x_i - 4\sin^3 x_i)$ which gives

$\sin x_i = 0$ or $\lambda-9=-12\sin^2 x_i$.

If $\sin x_i = 0$ then the corresponding $\cos 3x_i$ is maximized when $\cos x_i = 1$.

If $\sin x_i \neq 0$ then all remaining $\sin^2 x_i$ must be identical. Then all remaining $\cos^2 x_i$ are also identical. Let $\mu=|\cos x_i|$. Then $0\leq \mu\leq 1$.

Let $q$ be the number of indices with $\cos x_i = -\mu$, let $n$ be the number of indices with $\cos x_i = \mu$, and $m$ be the number of indices with $\cos x_i=1$.

Then necessarily, $q+n+m=10$ and the constraint becomes $$ m+(n-q)\mu=0. $$ The objective function becomes $$ \begin{align} m+ & n(4\mu^3-3\mu) + q(4(-\mu)^3 - 3(-\mu)) \\ &= m+ 4(n-q) \mu^3 - 3(n-q)\mu\\ &=4m+4(n-q)\mu \mu^2\\ &=4m(1-\mu^2)\\ &=4m\left( 1-\frac{m^2}{(q-n)^2}\right).\end{align} $$

This is maximized with $q=7$, $n=0$, $m=3$. When this happens, we have $7$ of $x_1, \ldots, x_{10}$ satisfying $\cos x_i= -\frac 37$ and the other $3$ of them have $\cos x_i=1$. The maximum is $$ 12\left(1-\frac{9}{49}\right)=\frac{480}{49}\approx 9.7959. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.