Question

please help me out with this one- a right circular cone has a surface area of 718 square inches. what dimensions (radius and height) will result in a maximum volume?

Answer 1

I began the problem, and now it after 30 mins of work it is a nightmare.

Answer 2

What class is this for--calculus?

Answer 3

yup

Answer 4

so far, i've taken the SA eqn and tried to optimize it with substitution, but then after finding a derivative it just is bad.

Answer 5

If you could find satellite of myininaya they could help you

Answer 6

this one, jwg lol

Answer 7

yep

Answer 8

is there a formula for the surface area of a rt circular cone?

Answer 9

SA=pir^2+pirl

Answer 10

and perferably one for volume as well to relate the 2?

Answer 11

1/3 something is all i can remember

Answer 12

1/3 volume of a cylinder perhaps?

Answer 13

V=-1/3pir^2h

Answer 14

Exactly

Answer 15

ok; looks like we are given an value for SA so that we can find an r perhaps?

Answer 16

The problem is the slant height adds a variable.

Answer 17

i take it thats the "l" part in the SA

Answer 18

yes

Answer 19

precisely

Answer 20

l = sqrt(h^2 + r^2) then

718 = pi r^2 + pi r(sqrt(h^2+r^2))

718/pir = r + (sqrt(h^2+r^2))

(718/pir) - r = sqrt(h^2+r^2)

[(718/pir) - r]^2 = h^2+r^2

[(718/pir) - r]^2 - r^2 = h^2

i wonder of thats going according to plan :)

Answer 21

at any rate we can define h in terms of r; aint pretty but doable

Answer 22

That should be progress.

Wonder where the asker is.

Answer 23

fetal position, sobbing maybe lol

Answer 24

V = 1/3 pi r^2 h

\[V=\frac{pi\ r^2\ \sqrt{(\frac{718}{pi\ r}-r)^2-r^2}}{3}\]
looks about right?

Answer 25

yes

Answer 26

Now I suppose we have to differentiate that and set to 0?

Answer 27

the 1/3 pi is a constant that can be put to the side and we are left with deriving:\[V=r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}\]

Answer 28

yep, find the zeros and the undefineds to test

Answer 29

\[V'=r'^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)'^{1/2}\]
\[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*((\frac{718}{pi\ r}-r)^2-r^2)'\]
\[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*(\frac{718}{pi\ r}-r)'^2-r'^2\]
\[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*2(\frac{718}{pi\ r}-r)(\frac{718}{pi\ r}-r)'-2r\]
yep, its a doozie; the wolf might be quicker

Answer 30

That's what I did.

Answer 31

..... slacker!! lol

Answer 32

proudly so

Answer 33

r=7.56

Answer 34

yeah, i was gonna say around 8

Answer 35

plug that value in to get the V max

Answer 36

I just ignored the miserable denominator and set the numerator to 0.

Answer 37

Must first find the height

Answer 38

[(718/pir) - r]^2 - r^2 = h^2

plug in yer r

Answer 39

yep

Answer 40

i dunno if solving in terms of h would have been easier

Answer 41

tooooooooooooooooooooooooooooo late

Answer 42

lol, since h was buried i assume not

Answer 43

h= 21.37

Answer 44

that should do it then

Answer 45

Except for checking to make sure those values result in the correct surface area.

Answer 46

it has to, we did everything correctly

Answer 47

Of course...no mistakes...ever!!

Answer 48

they wanted max volume

Answer 49

the wolf helped so we can always pass the blame ;)

Answer 50

working on that.

Answer 51

1279.02= max volume

Answer 52

Good to have a scapegoat.

Answer 53

Hey!! It's right. SA = 717.9758

Answer 54

Thank you very much.