Polynomial optimization problem

Art of Problem Solving

AoPS Online

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

High School Math Grades 9-12, Ages 13-18

Grades 9-12, Ages 13-18

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Polynomial optimization problem

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

ReticulatedPython

562 posts

ReticulatedPython

#1 h Mar 31, 2025, 2:51 PM

Y by

Let $p(x)=-ax^4+x^3$ , where $a$ is a real number. Prove that for all positive $a$ , $p(x) \le \frac{27}{256a^3}.$

This post has been edited 1 time. Last edited by ReticulatedPython, Mar 31, 2025, 3:03 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1384 posts

Lankou

#2 h Mar 31, 2025, 7:09 PM

Y by

The function has four real zeros: $0$ of multiplicity $3$ and $\frac{1}{a}$ of multiplicity $1$ . The extremities of the graph both approache the $-\infty$ so the function has an absolute maximum between $0$ and $\frac{1}{a}$
$p'(x)=x^3(-ax+1)$
The graph has a turning point at $x=\frac {3}{4a}$
At this point the maximum is $=\frac {27}{256a^3}$
Hence $p(x) \le \frac{27}{256a^3}.$

This post has been edited 1 time. Last edited by Lankou, Mar 31, 2025, 7:11 PM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

812 posts

#4 h Apr 2, 2025, 12:38 PM

Y by

Taking $a \in \mathbb{R}^{+}$ , the derivative of $p(x)$ equal to zero yields the critical values:

$p'(x) = -4ax^3+3x^2 = x^2(3-4ax) = 0 \Rightarrow x = 0, \frac{3}{4a}$ (i).

A second derivative check of (i) shows that:

$p''(x) = -12ax^2+6x \Rightarrow p''(0) = 0$ (an inflection point) and $p''\left(\frac{3}{4a}\right) = -\frac{9}{4a} < 0$ (a global maximum).

Thus, $p_{MAX} = p\left(\frac{3}{4a}\right) = \textcolor{red}{\frac{27}{256a^3}} \le \frac{27}{256a^3}$ .

QED

This post has been edited 2 times. Last edited by Mathzeus1024, Apr 2, 2025, 12:43 PM

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a