Observatory ?

R

Ryan Rigdon

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Jun 10, 2010
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here is my problem:

An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the hemispherical dome cost 7 times as much per square foot as the cylindrical wall, where are the most economic dimensions for a volume of 16000 cubic feet?

This is all i have so far V = ?r^2h Any help would be great.
 
Here's a start.

The volume of the cylinder is \(\displaystyle {\pi}r^{2}h\)

The volume of the hemisphere is \(\displaystyle \frac{2}{3}{\pi}r^{3}\)

Total volume: \(\displaystyle {\pi}r^{2}h+\frac{2}{3}{\pi}r^{3}=16000\)

Surface area of cylinder: \(\displaystyle \overbrace{2{\pi}rh}^{\text{body}}+\underbrace{{\pi}r^{2}}_{\text{base}}\)

Surface area of hemisphere: \(\displaystyle 2{\pi}r^{2}\)

Total surface area: \(\displaystyle 2{\pi}rh+3{\pi}r^{2}\)

Now, they want you to minimize surface area. The cost constraints must be implemented. Can you do that OK?.
 
Although it is not explicit in the problem, I don't think the base of the cylinder(floor) should be included in the surface (walls) calculation.
 
my work thus far

Given V: ?r^2h + 2/3?r^3 = 16000

Solve for h:

?r^2h = 16000 -2/3?r^3

h = 16000/(?r^2) - 2/(3)*r

Let k be the cost per square foot of the cylindrical wall. Add total surface are. The cost is

C =k(2?rh) + 7k(3?r^2) sub h in


= k(2?r(16000/(?r^2) - 2/3*r) + 7(3?r^2))

= k((32000/r - 4/3?r^2) + 21?r^2))

= k(32000/r + 59/3 ?r^2)

C' = k(-32000/(r^2) + 118/3 ?r)

Set C' = 0 solve for r^3

k(-32000/(r^2) + 118/3 ?r) = 0

-32000/(r^2) + 118/3 ?r = 0

-96000 + 118?r^3 = 0

118?r^3 = 96000

r^3 = 48000/(59?) ; r = (48000/(59?))^(1/3)


im stuck here

my next step i think would be to plug r into h

h = 16000/(?(48000/(59?))^(2/3)) - 2/3(48000/(59?))^(1/3)
 
Ryan Rigdon said:
my work thus far

Given V: ?r^2h + 2/3?r^3 = 16000

Solve for h:

?r^2h = 16000 -2/3?r^3

h = 16000/(?r^2) - 2/(3)*r

Let k be the cost per square foot of the cylindrical wall. Add total surface are. The cost is

C =k(2?rh) + 7k(3?r^2) sub h in <<< Why does your flooring (base of the cylinder) cost as much as hemispherical wall?


= k(2?r(16000/(?r^2) - 2/3*r) + 7(3?r^2))

= k((32000/r - 4/3?r^2) + 21?r^2))

= k(32000/r + 59/3 ?r^2)

C' = k(-32000/(r^2) + 118/3 ?r)

Set C' = 0 solve for r^3

k(-32000/(r^2) + 118/3 ?r) = 0

-32000/(r^2) + 118/3 ?r = 0

-96000 + 118?r^3 = 0

118?r^3 = 96000

r^3 = 48000/(59?) ; r = (48000/(59?))^(1/3)


im stuck here <<< Why? Just use calculator and find the numeric value.

my next step i think would be to plug r into h

h = 16000/(?(48000/(59?))^(2/3)) - 2/3(48000/(59?))^(1/3)<<< Just use calculator and find the numeric value.
Click to expand...
 
i was getting ready to type my answer and then the power went out in my apartment. hopefully it doesnt do it again. so all the work i did was for nothing because i have to do it again, same problem different digits. here is the new one i got when i signed back in

An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the hemispherical dome cost 7 times as much per square foot as the cylindrical wall, where are the most economic dimensions for a volume of 10000 cubic feet?


will be back shortly with my work again. keep your fingers crossed that it doesnt happen again with the power issue.
 
An observatory is to be in the form of a right circular cylinder surmounted by a hemispherical dome. If the hemispherical dome cost 7 times as much per square foot as the cylindrical wall, where are the most economic dimensions for a volume of 10000 cubic feet?

The volume of the cylinder is \(\displaystyle {\pi}r^{2}h\)

The volume of the hemisphere is \(\displaystyle \frac{2}{3}{\pi}r^{3}\)

Total volume: \(\displaystyle {\pi}r^{2}h+\frac{2}{3}{\pi}r^{3}=10000\)

Surface area of cylinder: \(\displaystyle \overbrace{2{\pi}rh}^{\text{body}}+\underbrace{{\pi}r^{2}}_{\text{base}}\)

Surface area of hemisphere: \(\displaystyle 2{\pi}r^{2}\)

Total surface area: \(\displaystyle 2{\pi}rh+3{\pi}r^{2}\)


My Work:

Given V: \(\displaystyle {\pi}r^{2}h+\frac{2}{3}{\pi}r^{3}=10000\)


Solve for h:

?r^(2) h = 10000 - 2/3 ?r^3

h = 10000/(?r^2) - 2/3 r

Let k be the cost per square foot of the cylindrical wall. Add total surface area. The cost is

C = k(2?rh) + 7k(3?r^2) Sub h in

= k(2?r(10000/(?r^2) - 2/3r) + 7(3?r^2))

= k((20000/r - 4/3?r^2) + 21?r^2)

= k(20000/r + 59/3 ?r^2)

C ' = k(-20000/(r^2) + 118/3 ?r) Set C ' = 0 then solve for r^3

-20000/(r^2) + 118/3 ?r = 0 -60000 + 118?r^3 = 0

r^3 = 60000/(118?) = 30000/(59?) , r = (30000/(59?))^(1/3)

plug r into h

h = 10000/(?(30000/(59?))^(2/3)) - 2/3(30000/(59?))^(1/3)

r is approx. 5.4

h is approx. 103.5


Thus my answer to the problem.

The radius of the cylindrical base(and of the hemisphere) is 5.4 ft. The height of the cylindrical base is 103.5 ft.


Look good? :D
 
Hey R man:

I think you may have multiplied the cost constraints by the incorrect surface area what nots.

I am going to include the base of the cylinder. A 'silo' with no base is not much good for anything, rather the problem states it or not. Hey, that's me :) . The problem does state 'compared to the cylinder wall, though.

The volume is \(\displaystyle {\pi}r^{2}h+\frac{2}{3}{\pi}r^{3}=10000\)

Solve this for h gives:

\(\displaystyle h=\frac{2(15000-{\pi}r^{3})}{3{\pi}r^{2}}\)

Sub this into the surface cost formula, \(\displaystyle S=7c(\underbrace{2{\pi}r^{2}}_{\text{dome}})+c(\overbrace{2{\pi}rh+{\pi}r^{2}}^{\text{cylinder w/ base}})\)

\(\displaystyle S=\frac{c(41{\pi}r^{3}+6000)}{3r}\)

Differentiate

\(\displaystyle S'(r)=\frac{2c(41{\pi}r^{3}-3000)}{3r^{2}}\)

Now, set this to 0 and solve for r.
 
\(\displaystyle h=\frac{2(15000-{\pi}r^{3})}{3{\pi}r^{2}}\)


set S' = 0


\(\displaystyle S'(r)=\frac{2c(41{\pi}r^{3}-3000)}{3r^{2}}\)

41?r^3 - 3000 = 0

r = (3000/(41?))^(1/3) plug r into h

h = (2(15000-?(3000/(41?))) / (3?(3000/(41?))^(2/3))


r is approx. 2.9 h is approx. 388.4



The radius of the cylindrical base(and of the hemisphere) is 2.9 ft. The height of the cylindrical base is 388.4 ft.


does this look correct for the final answer.
 
I am sorry, Ryan. I left out a 0.

The surface should be

\(\displaystyle S=\frac{c(41{\pi}r^{3}+60000)}{3r}\)

\(\displaystyle S'=\frac{2c(41{\pi}r^{3}-30000)}{3r^{2}}\)

I get r=6.15 and h=79.98

Remember, I used the base. This may not be what they were getting at. But, to me, the base should be included.
 
when i recalculated r is approx. 6.2 h is approx. 80 we were told to round to the nearest tenth.



i am tired of looking at this problem and the next one is just as hard. Crap!!!!
 
Post the next one. We welcome challenges and difficult problems. You show lots of effort, so we do not mind helping you along.

The tough ones are the ones that make us better. If they were all easy, we would learn very little.
 
so is it safe to say that when i recalculated r is approx. 6.2 h is approx. 80 is the correct answer. :shock:
 
That's what I got. But remember, I used the base. Maybe they were getting at something else like SK mentioned.
 
i am going to leave it. posted my other problem. any helpful hints would be great. my brain if fried.
 
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