Question

Here is a ridiculously easy sample question from Oklahoma University's document on preparing for the Putnam Exam. Whoever solves first gets a medal. Took me about 3-4 minutes without a calculator =P. If you beat me on the time as well, then I'll fan you too :D

"Find, with explanation, the maximum value of \(\bf f(x)=x^3-3x\) on the set of all real numbers \(\bf x\) satisfying \(\bf x^4+36 \le 13x^2\)."

Answer 1

4:31 AM right now. Your time begins now. (You may also time yourself independently and please be honest lol)

Answer 2

Also, please try to solve the question YOURSELF, i.e. without getting help from ANY other source.

Answer 3

this is what you do for fun?

Answer 4

lol

Answer 5

umm guess its 18 since domain is 2<x<3

Answer 6

LOL @dumbcow got it. but he gets no medal since he guessed.

Answer 7

haha :)

Answer 8

had he not included the word "guess" and i would've given him half a medal =]

Answer 9

I should also include, legit solutions only.

Answer 10

we know it has a max, but -1 and 1 does not satisfy the restraint, so solve the restraint and check the endpoints of the interval.

Answer 11

@zzr0ck3r will provide that for me

Answer 12

im not good at solving those things

Answer 13

lol

Answer 14

@zzr0ck3r you're about to be a math major. stop kidding me.

Answer 15

tell someone to let a = x^2 and that might help

Answer 16

and for the odd powers, i don't think the squared substitution would help

Answer 17

@zzr0ck3r did make a good suggestion which can be used in SOME part of the question which I will not state.

Answer 18

basically re-arrange the inequality, so it comes down to finding the maximum of a function over some interval

Answer 19

to lazy to work out details

Answer 20

Eh, you find the domain that was mentioned previously, in which you check your interval endpoints, get x = 3, plug it in and get 18 as the max value.

Answer 21

thats what i said

Answer 22

well looking at the inequality you get

\[(x^2 -9)(x^2 -4)\le 0\]

Answer 23

@Jack111 Everyone is telling me random stuff but noone has given a solution.

Answer 24

I know :P

Answer 25

x^4 +36-13x^2≤0
(x^2)^2-13(x^2)+36<=0
(x^2-9)(x^2-4)<=0

Answer 26

check 3,-3, 2, -2. one will be 18

Answer 27

i am to lazy, if you know pretty much have the idea and the rest is just computation, then why bother

Answer 28

if @dumbcow was right

Answer 29

@zzr0ck3r stop helping unless you're ACTUALLY going to finish the solution lol.

Answer 30

I think you can do this with Lagrange multipliers as well..

Answer 31

The beauty of problem-solving is solving questions creatively with limited knowledge. That's how we gain and create more mathematical knowledge. In short, there is no need to use lagrange multipliers for a problem as simple as this.

Answer 32

we know it has a max
y'=0 gives x=-1,1 and these do not satisfy the restraint
x^4 +36-13x^2≤0
(x^2)^2-13(x^2)+36<=0
(x^2-9)(x^2-4)<=0
this is cubic so it will be the "right end of the interval"
f(3) = 18

Answer 33

cubic with positive coeffecient

Answer 34

ps of course I know there is no need:) and its pretty much the same thing anyway

Answer 35

@zzr0ck3r You're solution is to an extent correct but still lacking.

Answer 36

lol how so?

Answer 37

and the domain would be [-3,-2] U [2,3]

Answer 38

OK, I feel its complete enough now lol. @zzr0ck3r wins.

Answer 39

I should have said its cubic with no square term

Answer 40

$$x^4-13x^2+36\le 0\\(x^2-9)(x^2-4)\le 0$$hence for \([-3,-2]\cup[2,3]\) is the domain of optimization. for \(f(x)=x^3-3x\) observe:$$0=3(x^2-1)\\f''(x)=6x$$ i.e. \(x=\pm 1\) are our relative extrema. for \(x=-1\) we see \(f''<0\) i.e. negative curvature so we have a maximum at \(x=-1\) but this is not in our domain of optimization. since \(f\) is increasing an unbounded for \(x\gt1\) the answer is \(x=3\)

Answer 41

not sure how long it took me probably 3-4 minutes

Answer 42

you still didn't give the max value:P

Answer 43

oh derp \(f(3)=3^3-3^2=18\)

Answer 44

:)