Question

Calc Optimization Help Please?

Answer 1

The sum of two positive numbers is 16. Find the possible max and min value of the sum of their cube roots.

Thnk you very much

Answer 2

let the 2 numbers be 'x' and 'y'
so, x+y=16
to maximize f(x) = x^3+y^3
can you find f(x) in terms of 'x' only ?

Answer 3

sorry, f(x) = \(\sqrt[3]x+\sqrt[3]y\)

Answer 4

Doesnt cube root mean , ahh ok :)

Answer 5

16 - x^1-3 = y

Answer 6

thats what happen when you read the question real fast :P

Answer 7

Haha, so y = (16 - X^(1/3))

Answer 8

and xy = a min or max, so there derivative of that = o?

Answer 9

from x+y=16
y is just 16 -x right ?

Answer 10

substitute this in f(x)

Answer 11

Oh yep

Answer 12

" xy = a min or max, " what ?

Answer 13

Whoops never mind

Answer 14

sorry sorry wrong idea

Answer 15

f(x) = x^(1/3) + (16 - x^3)^1/3

Answer 16

so, f(x) =...?

Answer 17

is it f(x) = x^(1/3) + (16 - x^3)^1/3?

Answer 18

did you mistype x^3 ?
it would only be 'x'

Answer 19

y^(1/3) ----->(16-x)^(1/3)

Answer 20

Yes sorry sorry

Answer 21

you are right

Answer 22

now to max/min , take derivative of f(x) and equate it to 0

Answer 23

Now I have (16 - x)^2/3 - x^2/3 = 0

Answer 24

hmm..did you miss -*minus* sign ?
1/3 -1 = ... ?

Answer 25

Oh I skipped some steps, after I dervived them I made them common denominator and got rid of the denominator, should I should my steps?

Answer 26

ohh..got it....not required.
can you find 'x' from there ?

Answer 27

is x = 8?

Answer 28

yes! good work!

Answer 29

Thank you :)) The answer does not match my teachers, but its not the first time she's wrong haha

Answer 30

so, when x=8, f(x)=...?

Answer 31

but thats not final answer

Answer 32

4?

Answer 33

4 is correct.
did it match ?

Answer 34

it says minimum = 16^1/3

Answer 35

oh but it also says maximum is =4, so that part is correct

Answer 36

yes, because 4 is maximum (do you know how ?)
and we are yet to find minimum

Answer 37

ohhh, can you please tell me what would I do differently to find the minimum?

Answer 38

I would set the derivative to zzero again, correct? but I would get the same answer, 8?

Answer 39

yes, for (16 - x)^2/3 - x^2/3 = 0
you'll get more values of 'x' including 8,
x^(2/3) = (16-x)^(2/3)
cubing both sides,
x^2 = (16-x)^2
whats values of 'x' you get here ?

Answer 40

for that its only 8, ...

Answer 41

yeah I only get 8

Answer 42

Is there any way I am supposed to get a different answer as well?

Answer 43

yes, it is...
what about the denominator [x(16-x)]^(2/3)

Answer 44

OHHH I can make it undefined??

Answer 45

THIS is what I have been doing wrong for the past questions

Answer 46

I thought I had to make it = 0, not UD

Answer 47

Thank you once again, I get it now :)

Answer 48

not making it undefined (have to make it 0 only ), but.....let me think of an explanation.

Answer 49

if the graph of y' is UD then it can be a min/max?

Answer 50

I will review my notes =)

Answer 51

ok, this can be it, for f(x), x cannot be < 0 or >16
so, we need to test f(x) at end-points also for possible minimas/maximas.

Answer 52

put, x=0, x=16 in f(x), you'll get a values lees than what you got for x=8
hence you conclude its a minima

Answer 53

mathematical way to find whether a point is minima or maxima is to find double derivative, f'' (x) at those points,
if f''(x) > 0, then its a minima point
if f''(x) < 0, then its a maxima point

Answer 54

thank you~~

Answer 55

\[x+y=16 \implies y = 16-x\]\[f(x)=\sqrt[3]{x}+\sqrt[3]{16-x}\]Maximum/Minimum occur where the first derivative is 0. So let's take the derivative of f(x).\[f'(x)=\frac{ 1 }{ 3 }x^{-2/3}+\frac{ 1 }{ 3 }(-1)(16-x)^{-2/3}=\frac{ 1 }{ 3\sqrt[3]{x^2} }-\frac{ 1 }{ 3\sqrt[3]{(16-x)^2} }\]\[f'(x)=\frac{ \sqrt[3]{(16-x)^2}-\sqrt[3]{x^2} }{ 3\sqrt[3]{x^2(16-x)^2} }\]\[\frac{ \sqrt[3]{(16-x)^2}-\sqrt[3]{x^2} }{ 3\sqrt[3]{x^2(16-x)^2} }=0 \rightarrow \sqrt[3]{(16-x)^2}-\sqrt[3]{x^2}=0 \rightarrow \sqrt[3]{(16-x)^2}=\sqrt[3]{x^2}\]\[\rightarrow 16-x=x \rightarrow 16= 2x \rightarrow x =8\]AND\[f'(x) :D.N.E \rightarrow 3\sqrt[3]{x^2(16-x)^2}=0 \rightarrow x^2(x^2-32x+256)=0\]\[\rightarrow x^2(x-16)^2 =0 \rightarrow x =0,16 \]So our Critical points are, x = 0, 8, 16. These are the points at which we check for local/absolute Maximum/Minimum. We can do this by evaluating the values of f(x) at each x-value (0,8, or 16) in the closed interval [0,16] and simply compare which one is higher and which is lower and that way we can tell if it's a maximum or a minimum.\[f(0)\approx 2.52\]\[f(8)=4\]\[f(16) \approx 2.52 \]Clearly, x = 8 is a maximum and the other two points are minima. So the the highest sum occurs at x = 8, which is y =4 and the lowest sum occurs at x = 0, 16 which is y = 2.52 (approximately). And we are done.

@pottersheep

Answer 56

@hartnn

Answer 57

critical point for f(x) is just x=8
nothing else.

Answer 58

critical point : f'(x) = DNE <-----NO

Answer 59

http://www.wolframalpha.com/input/?i=critical+points+x%5E%281%2F3%29%2B+%2816-x%29%5E%281%2F3%29

Answer 60

You are wrong. Critical points are values where the derivative is 0 or does not exist.

Answer 61

Get your facts straight before calling someone wrong.

Answer 62

@agent0smith
Just go over my solution. It tells you everything. And no, we have to check whether it's zero/non-differentiable, and then we have to check whether these points where the function is zero/non-differentiable is a minimum or maximum.

Answer 63

@genius12 , sorry, i didn't mean to call you wrong, you are right, i need to revise my calc notes.

Answer 64

It's ok. All good =]

Answer 65

Thank you both of you, lots of help