Given $c > 0$ and $u > l > 0$, $$ \max_{x \in \Bbb R} \, \left( 1 - 2 c x^2 \right)^2 \quad \text{subject to} \quad l \leq x \leq u $$ Can the maximum value be found in terms of $l$ or $u$?

My try:

One can write the problem as follows: $$ \min_{ \begin{aligned} x&> 0\\ x- l &\geq 0\\ u -x &\geq 0 \end{aligned} } -(1-2cx^2)^2 $$ and define Lagrange multiplier $\lambda_1, \lambda_2, \lambda_3 \geq 0$ and consider 8 different situations for $\lambda$'s.


Is there a hacky way of solving the above without considering constrained optimization?

  • $\begingroup$ You mean you want to find the value of c that optimizes the value of the polynomial? $\endgroup$
    – MSIS
    Commented Nov 26, 2022 at 21:14
  • $\begingroup$ Consider the variations of $1-2cx^2$. When it's positive the max of the square is the max of $1-2cx^2$. When it's negative you will have to find the minimum. First do that without constraints. Then either the optimum is inside $(l,u)$ and then the constraints are not important, or on the border, and then depending on a condition on $c$ it's either $l$ or $u$. $\endgroup$ Commented Nov 26, 2022 at 21:33

1 Answer 1


Using the constraint $0<l≤x≤u$, we have:


This implies that,

$$ \begin{align}\max \left\{\left(1-2cx^2\right)^2\wedge 0<l≤x≤u\right\}=\max\left\{\left(1-2cu^2\right)^2,\;\left(1-2cl^2\right)^2\right\}\end{align} $$

  • 1
    $\begingroup$ +1 Of course. I forgot the "$0<$" in $0<l\le x\le u$, an this makes things much simpler as there is no local max. $\endgroup$ Commented Nov 26, 2022 at 21:38

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