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?

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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?

Mathematics
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I began the problem, and now it after 30 mins of work it is a nightmare.
What class is this for--calculus?
yup

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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.
If you could find satellite of myininaya they could help you
this one, jwg lol
yep
is there a formula for the surface area of a rt circular cone?
SA=pir^2+pirl
and perferably one for volume as well to relate the 2?
1/3 something is all i can remember
1/3 volume of a cylinder perhaps?
V=-1/3pir^2h
Exactly
ok; looks like we are given an value for SA so that we can find an r perhaps?
The problem is the slant height adds a variable.
i take it thats the "l" part in the SA
yes
|dw:1327287244737:dw|
precisely
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 :)
at any rate we can define h in terms of r; aint pretty but doable
That should be progress. Wonder where the asker is.
fetal position, sobbing maybe lol
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?
yes
Now I suppose we have to differentiate that and set to 0?
the 1/3 pi is a constant that can be put to the side and we are left with deriving:\[V=r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}\]
yep, find the zeros and the undefineds to test
\[V'=r'^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)'^{1/2}\] \[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*((\frac{718}{pi\ r}-r)^2-r^2)'\] \[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*(\frac{718}{pi\ r}-r)'^2-r'^2\] \[V'=2r\ ((\frac{718}{pi\ r}-r)^2-r^2)^{1/2}+\frac{1}{2}r^2\ ((\frac{718}{pi\ r}-r)^2-r^2)^{-1/2}*2(\frac{718}{pi\ r}-r)(\frac{718}{pi\ r}-r)'-2r\] yep, its a doozie; the wolf might be quicker
That's what I did.
..... slacker!! lol
proudly so
r=7.56
yeah, i was gonna say around 8
plug that value in to get the V max
I just ignored the miserable denominator and set the numerator to 0.
Must first find the height
[(718/pir) - r]^2 - r^2 = h^2 plug in yer r
yep
i dunno if solving in terms of h would have been easier
tooooooooooooooooooooooooooooo late
lol, since h was buried i assume not
h= 21.37
that should do it then
Except for checking to make sure those values result in the correct surface area.
it has to, we did everything correctly
Of course...no mistakes...ever!!
they wanted max volume
the wolf helped so we can always pass the blame ;)
working on that.
1279.02= max volume
Good to have a scapegoat.
Hey!! It's right. SA = 717.9758
Thank you very much.

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