Question

Find the maximum value of f(x) = x³ - 6ax² + 9a²x - 2a³ in the interval 00. I will draw my work, pls help

Answer 1

Actually, the interval is \[0 \le x \le 2\]

Answer 2

I did this: f'(x) = 3x² - 12ax + 9a² => 3(x² - 4ax + 3a²) => 3(x-3a)(x-a)

Answer 3

Looks good so far.

Answer 4

K, I also did a table.

Answer 5

From there, recall that local maximums and minimums are at the zero's of the derivative. So the next step would be to set your derivative equal to 0.

Answer 6

Check this

Answer 7

I don't know if it's understandable lol

Answer 8

What are the possible solutions for x to \[3(x-3a)(x-a)=0\;?\]

Answer 9

Those I draw. 0, a and 3a.

Answer 10

Ah, I see what you had written down now. First, note that 0 can't be a solution unless \(a\) was also 0, but \(a>0\) so 0 isn't a solution. So using your solutions of \[x=a\]\[x=3a\]plug those solutions back into the original equation for x.

Answer 11

In f'(x) or f(x)?

Answer 12

f(x), since we're looking for maximum of f(x).

Answer 13

I already draw that. When x =a, f(x) = 2a³. When x = 3a, f(x) =0. Check the table.

Answer 14

I also did it with f'(x) to see the negatives and positives

Answer 15

Well, since \(a>0\) we know that \(2a^3>0\) so \(x=2a^3\) is a maximum.

Answer 16

I have to know the maximum when a>2

Answer 17

I guess I need to draw the graph, pellet :/

Answer 18

why when \(a>2\)? Isn't it in the interval \(0\leq x\leq2\)?

Answer 19

you should find y'' so you can decide if x=a or x=3a are max or mins.

Answer 20

Yea, but to find the values in that interval I have to first find when a>2 and all those others. I can't explain it without the graph

Answer 21

What phi said is actually a great way to do it. Find the second derivative, and then plug in the values of x where the first derivative was 0. If it's negative, then you have a maximum, and if it's positive you have a minimum.

Answer 22

I know that, but the problem is the interval

Answer 23

The example says I should do the graph

Answer 24

I think I see the issue. You also need to check what happens when \(x=0, 2\) because those might be the max and min in the interval.

Answer 25

Yea

Answer 26

If you let x=0, you get \(f(0)=-2a^3\) and since \(a>0\), \(a^3>0\) so \(f(0)\) can't be a maximum.

Answer 27

You're not understanding what I mean, actually. Check my table. I already did that, now I need to find the values in those intervals

Answer 28

I hate this exercise lol

Answer 29

Well, If you try it with \(x=2\), you find \(f(2)=8-24a+18a^2-2a^3\).

Answer 30

Quick question, is \(a>0\) or \(a>2\)?

Answer 31

a>0. But in the example a was also > 0 and they solved using a>2, a<2<4a and 4a<2, and the found values were like f(2), f(a), etc

Answer 32

break the problem into
0 < a ≤ 2 where f(a) will be a maximum
2<a where f(2) will be the maximum

Answer 33

I tried to take a pic of the example but my cam sucks lol

Answer 34

How do I do that, phi?

Answer 35

Check those graphs and the alternate places where I have to put x so I can find the values in the interval

Answer 36

King, you there?

Answer 37

Notice that the first graph is for a>2

Answer 38

You already have (almost)
break the problem into
0 < a ≤ 2 plug in a into f(x) that is the answer
2<a where f(2) will be the maximum

Answer 39

In the chart, the top row is x, second row is f'(x), and third row is f(x) correct?

Answer 40

King already evaluated f(2) up above

Answer 41

Yes, King

Answer 42

I can't see that, phi

Answer 43

I guess I already did that

Answer 44

lol i'm confused now

Answer 45

Think of it this way. If a is less than 2, where is the maximum of the function going to be

Answer 46

You know f(a) is a maximum.

Answer 47

On the other hand, if a>2
the function is rising toward f(a)
so f(2) (the boundary of your interval) must the max

Answer 48

Oh, that's right!

Answer 49

But what about the others?

Answer 50

The question is

Find the maximum value of f(x) = x³ - 6ax² + 9a²x - 2a³ in the interval 0<x<2. Assume a>0.

You did:

0<a≤2 2a^3
2<a -2a^3+18a^2-24a+8

Answer 51

But you're confusing the intervals. When you draw the graph and you alternate positions of a, you have to find the values in those intervals as well

Answer 52

I will give the example. Let f(x) be x³ - 6ax² + 9a²

Answer 53

for the same interval

Answer 54

and a>0 as well

Answer 55

It would go like:

when a>2, max is f(2) = 18a² -24a + 8
when a \[\le\] 2 \[\le\] 4a, that is, 1/2 \[\le\] a \[\le\] 2 (notice both intervals) then max is f(a) = 4a³
when 4a<2, that is, 0<a<1/2, max is f(2) again

Answer 56

I meant \[a \le 2 \le 4a\] and \[1/2 \le a \le 2\]

Answer 57

in this case f(3a) is a minimum, f(a) is a max. a>0 (so f(0) can't be the max)

Answer 58

I think finally see what's going on now. And I agree with phi. We have to break the function up into \(0<a\leq2\) and \(2<a\). Then, we can find the maximum of the function in those parameters.

Answer 59

Problem is I can't see that without graphs

Answer 60

OK, I think I see what you are saying