Question

Optimization. Please check and help! :)

Answer 1

Original Question

Answer 2

My work so far

Answer 3

@zepdrix free? :)

Answer 4

Ok cool.
So you've left off at part c?
Trying to get P(r), yes?

Answer 5

Yeah I'm not sure where to go now with it. I would have to move stuff over from the right and divide p by it, yeah?
end up with p/(something) on the left hand side?

Answer 6

no no no.
Right now you have P(r,h).
See how your cost function is written in terms of r and h stuff?
We need to get rid of the h's.

We need some way to replace h's with a r stuff,
that will turn P(r,h) into P(r).

Answer 7

We will do that by using the volume function.
That first ... half of a sentence that you wrote in part c is really important.

Answer 8

V is no longer a variable,
we're holding volume constant, plugging in some value for v.

Answer 9

ooo. I see o.0

Answer 10

Wait but there isn't even a v in the p = (stuff)
Why does it matter that v is a constant there?

Answer 11

oh! we plug the V= into the P= ?

Answer 12

Your V formula, shows a relationship between r and h.
I should say, that it shows a relationship between V, r, and h.
But if V is constant, then we only have 2 variables!
The ones that we want! :)

So plug in a constant for V, maybe v_o or C or something to remind your self that it's no longer constant.

And solve for h in that formula.

Answer 13

to remind yourself that it's no longer `a variable`*
typo

Answer 14

hmm I see. So all that stuff under C i should just erase xD

Answer 15

and start with writing V as h = (Stuff)

Answer 16

No everything under C looks great so far.
After that, you'll need to replace the h in P with ...
whatever you get from solving for h in your V formula.

Answer 17

I mean, the order doesn't really matter :)
You don't need to START with V=(h stuff),
unless you prefer to organize it that way.

Answer 18

Nah i'll just solve it off to the side hehe
does h = v-(4/3)r ?

Answer 19

Grr I'm on my laptop with no lights on XD
No paper around

Hard to check your work haha.
But I think... you forgot to divide v by the pi r^2.

Answer 20

\[V=\frac{ 4 }{ 3 }\pi r^3+h \pi r^2\]
\[V-\frac{ 4 }{ 3 }\pi r^3=h \pi r^2\]
\[h=\frac{ v-(4/3) \pi r^3 }{ \pi r^2 }\]
The pi and r^2 divide out

Answer 21

No, it only divides out of the second part :)

Answer 22

\[\large\rm h=\frac{v}{\pi r^2}-\frac{\frac{4}{3}\pi r^3}{\pi r^2}\]

Answer 23

oh I see!
is it simpler to keep it like that or shall I go ahead and divide it out of the second part?

Answer 24

h = (v/pi r^2) - (4/3)r

Answer 25

Hmm, it's hard to kind of look ahead and see which will be more beneficial.
But I'm "thinking" that this is probably the best form to use:\[\large\rm h=\frac{1}{\pi r^2}\left(v-\frac{4}{3}\pi r^3\right)\]I could be wrong, yours might be better.
I'm just thinking about how things might cancel out when we finally plug this in.

Answer 26

hmm ok

Answer 27

I guess it doesn't really make a difference which way xD then I just plug that into where I left off with p = ?

Answer 28

ya

Answer 29

yaay. And that finishes c? :D

Answer 30

yup sounds good \c:/

Answer 31

Part d I don't even know where to start.

Answer 32

I don't see a part d anywhere 0_o
I only see 1-7

Answer 33

oh! sorry! just a moment.

Answer 34

Ummm well,
what happens when your radius is length 0?
What will your price be?

Answer 35

0!

Answer 36

If I were to give the entire thing the same denominator the denominator would be 3(pi(r)^2)
so r = 0 is the only solution?

Answer 37

The domain is an interval :)

Clearly r=0 is the minimum length,
we can't have negative length.

I'm just trying to think about whether or not we would have a maximum r we can plug in.

Answer 38

Oh, and perhaps we can't even have 0,
did you end up with an r in your denominator of your p function somewhere? :o

Answer 39

Something like this? :o\[\large\rm p(r)=\frac{8k \pi r^3+2k^2(v-\frac{4}{3}\pi r^3)}{r}\]

Answer 40

Eem I didn't simplify heh

Answer 41

I probably should then xD
Well if you look at the function right after plugging h in then we can see that the denominator would be 3(pi r^2)

Answer 42

Oh I left the 3 up top :d
But ya you could do that.

Answer 43

Domain shouldn't be too difficult, the radius can be as small as 0, but not including 0, and ... as large as we want, i think..

\(\large\rm r\in (0,\infty)\)
something like that prolly? :o

Answer 44

looks good! I shall double check tomorrow with the prof but I think that is fine :D

Answer 45

Part e!

Answer 46

So this is the fun calculus part of the problem :)
Take your derivative, do the stuff, find the critical points

Answer 47

woo. the derivative of function p(r) yeah?

Answer 48

ya :d

Answer 49

Gwoow it's so long though haha. This will be an adventure! So probably here I should actually simplify

Answer 50

and give it all a common denominator etc etc.

Answer 51

i would ummm...

Answer 52

I would split it into three terms.
The denominator divides out of two of the terms.
Put a negative power on the one that doesn't.

And then apply power rule to all three.
Maybe take that route.. I dunno

Answer 53

how did you get the 3 up top?

Answer 54

I dunnoooo >.<
I didn't check your p function,
So I dunno if it's right lol

But if it is right,
then plugging in your h,\[\large\rm P(r)=8k \pi r^2+2\pi k^2 r h\]\[\large\rm P(r)=8k \pi r^2+2\pi k^2 r \cdot\frac{1}{\pi r^2}\left(v-\frac{4}{3}\pi r^3\right)\]Distributing the stuff to the brackets gives you three total terms, yes?

Answer 55

\[\large\rm P(r)=8k \pi r^2+2v k^2 r^{-1}-\frac{4}{3}\pi r\](If I did that correctly)

Answer 56

No I did the last term incorrectly, woops

Answer 57

\[\large\rm P(r)=8k \pi r^2+2v k^2 r^{-1}-\frac{4}{3}k^2 \pi r^2\]

Answer 58

And then just power rule power rule power rule, ya? :o

Answer 59

mm....just a sec double checking

Answer 60

on the last term wouldn't it be 8 rather than 4 because you multiply that 2 in?

Answer 61

Sideways, sorry!

Answer 62

oh wait. i think mine is wrong because i would have pi^2 too

Answer 63

ahh the 8, yes

Answer 64

Ahhh I dunno,
my brain is fried, I need a math breeeakkk >.<
Sorryyyyyy! maybe tomorrow XD

Answer 65

Awe ok! no worries :)
go to sleep xD

Answer 66

:D

Answer 67

I shall try to finish it and then if you have recovered by tomorrow I shall upload a finished pic :)

Answer 68

Original Question!

Answer 69

My work! I think I included everything necessary. Sorry for so many pictures xD they are in order and these aren't sideways haha

Answer 70

@Kainui I know this is quite long. Would you be willing to check my work? :)
Everything is included in these last two posts by me. (so you don't have to go looking through the whole chain). I only post it in this chain to keep it organized for myself.

Answer 71

@zepdrix you are back! Didn't scare you away, eh? I think I finished solving the entire thing (through there may be errors) I got rid of a ton of errors that I found this morning from last night's work o.o

Answer 72

No more k^2?
Ah good good. I was wondering about that :p

Answer 73

yeeeah. I had accidentally included the circles on the top and bottom of the cylinder.

Answer 74

Ohhh

Answer 75

yep

Answer 76

Grr I keep making boo boos when I'm checking it :P

Answer 77

Take your time ^-^

Answer 78

Woops, small typo on your critical point.
It magically turned from a 3 to a 10 in the numerator.

Answer 79

Christmas magic. o.o

Answer 80

Gwow Idk how that happened haha =.=

Answer 81

So is there any way to check if this is really the min? Because of that V I don't know how.

Answer 82

Ummm

Answer 83

Well, it's a volume, so we know that V>0.
That should allow us to do the "test point" type thing,
I think,

assuming the k's don't cause a problem.

Answer 84

Oh but k is also positive, it's a price, right?

Answer 85

Well at that point the k's aren't even there.
Yep.
Also the whole thing on the right side of r = should have a +/- in front of it but it has to be positive, right?

Answer 86

Well I was talking about P',
which does involve some k's.
But umm

Answer 87

Oh I see. Well like to perform the first derivative test on the r..

Answer 88

I don't know how you would come up with test points larger or smaller than \(\large\rm \sqrt[3]{\frac{3v}{10\pi}}\)

I guess you could drop the 3 under the root and that makes it smaller,\[\large\rm x=\sqrt[3]{\frac{v}{10\pi}}\]^That would be a point on the left side of your critical value.
You could plug it in, and maybe come up with a sign.

Sounds so tedious though XD hah

Answer 89

Gross haha eem..maybe it's fine without it xD

Answer 90

Seems like everything is in order
Yay good job nini! \c:/

Answer 91

Thank you!!