Optimization:container with a hemisphere ontop of a cylinder

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wind

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A closed container is made with a hemisphere on top of a cylinder. the height and the radius of the cylinder are h and r respectively. given that the surface area of the container is 20cm^2 fond all dimensions of the container (the radius and height) that will maximize the volume if the container.

Sphere
S= 4pir²
V= 4/3pir³

Cylinder
V= pir²h
S= 2pirh + 2pir²

Surface area of the conaitner

20=2pirh + 2pir² + (4pir²)/2
20=2pirh + 4pir²
(20-4pir²)/2pir=h

Total Volume

V= 4/3pir³ + pir²h
V= 4/3pir³ + pir²((20-4pir²)/2pir)
V= 4/3pir³ + (20pir²-4pir²)/2pir
V= 4/3pir³ + 2pir(10r-2r)/2pir
V= 4/3pir³ + (10r-2r)
V= 4/3pir³ + 8r

Find v'

V'= 4pir² + 8

Maximize

0 = 4pir² + 8
-8 = 4pir²
root(-8/4pi) =r

Did I make a mistake somewhere?
 
Re: Optimization:container with a hemisphere ontop of a cyli

wind said:
A closed container is made with a hemisphere on top of a cylinder. the height and the radius of the cylinder are h and r respectively. given that the surface area of the container is 20cm^2 fond all dimensions of the container (the radius and height) that will maximize the volume if the container.

Sphere
S= 4pir²
V= 4/3pir³

Cylinder
V= pir²h
S= 2pirh + 2pir²
There is only one circular face.

Surface area of the conaitner

20=2pirh + 2pir² + (4pir²)/2
20=2pirh + 4pir²
(20-4pir²)/2pir=h

Total Volume

V= 4/3pir³ + pir²h
The volume of the >hemi<sphere is?
V= 4/3pir³ + pir²((20-4pir²)/2pir)
V= 4/3pir³ + (20pir²-4pir²)/2pir
4pir^2 * pir^2 = ?
V= 4/3pir³ + 2pir(10r-2r)/2pir
V= 4/3pir³ + (10r-2r)
V= 4/3pir³ + 8r

Find v'

V'= 4pir² + 8

Maximize

0 = 4pir² + 8
-8 = 4pir²
root(-8/4pi) =r

Did I make a mistake somewhere?
Click to expand...
 
Be careful. You have a hemisphere, not a full sphere. Therefore it's volume is \(\displaystyle \L\\\frac{2}{3}{\pi}r^{3}\)

Surface of hemisphere is \(\displaystyle \L\\2{\pi}r^{2}\)


\(\displaystyle \L\\S=3{\pi}r^{2}+2{\pi}rh=20\)

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

Now, I assume you know what to do?.
 
A closed container is made with a hemisphere on top of a cylinder. the height and the radius of the cylinder are h and r respectively. given that the surface area of the container is 20cm^2 fond all dimensions of the container (the radius and height) that will maximize the volume if the container.

Sphere
S= 4pir²
V= 4/3pir³

Cylinder
V= pir²h
S= 2pirh + 2pir²

Surface area of the conaitner

20=2pirh + 2pir² + (4pir²)/2
Click to expand...

Not correct; the hemispherical top replaces one of the circular ends of the cylinder:

20=2pirh + pir² + (4pir²)/2
20=2pirh + 3pir²

Total Volume

V= 4/3pir³ + pir²h
Click to expand...

Also incorrect. A hemisphere is only half a sphere:

V= 2/3pir³ + pir²h


Hope that helps.
OOps. I see Galactus was too fast for me. :) And Unco too!
 
Thanks, Unco,Galactus, wjm11 :D

But I'm still stuck... :?

Surface Area
20=2pirh + 3pir²
(20- 3pir²)/2pir=h

Volume

V= 2/3pir³ + pir²h
V= 2/3pir³ + pir²((20- 3pir²)/2pir)
V= 2/3pir³ + (20pir²- 3pir²)/2pir
V= 2/3pir³ + pir(20r- 3r)/2pir
V= 2/3pir³ + (20r- 3r)/2
V= 2/3pir³ + 17r/2

v'= 2pir² + 17/2

but if i were to solve for r, there is a negative under the root?

Am I doing something wrong? Thanks
 
In my first reply I pointed out your error in multiplying 4pir^2 and pir^2. You have made the same mistake.
 
^oh thanks Unco , I missed that


Volume

V= 2/3pir³ + pir²h
V= 2/3pir³ + pir²((20- 3pir²)/2pir)
V= 2/3pir³ + (20pir²- 3pi²r^4)/2pir
V= 2/3pir³ + 10r- 3/2pir³

V'= 2pir² + 10- 9/2pir²
V'= -5/2pir² + 10

0= -5/2pir² + 10
-10= -5/2pir²
1.128≈r

Find h
just plug the value for r into this euqation right?
(20- 3pir²)/2pir=h

Is this right?
 
You got it. :D

Except, maybe leave your answer in the form \(\displaystyle \L\\r=\frac{2}{\sqrt{\pi}}\). Looks better.
 
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