# Maximizing a quartic polynomial over an interval

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.

Question

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

• You mean you want to find the value of c that optimizes the value of the polynomial?
– MSIS
Commented Nov 26, 2022 at 21:14
• 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 $u$. Commented Nov 26, 2022 at 21:33

Using the constraint $$0, we have:
$$1-2cu^2≤1-2cx^2≤1-2cl^2$$
\begin{align}\max \left\{\left(1-2cx^2\right)^2\wedge 0
• +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. Commented Nov 26, 2022 at 21:38