Could you please check my working and help me complete..thank you

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Apr 30th 2008, 09:25 AM
Chez_

Could you please check my working and help me complete..thank you

I have a q..bin with volume pie/8m^3 made from metal, to keep costs low it should have the min possible surface area, Am^2.
* using pie/8m^3 write an expression for height t in terms of radius r..
Which iv done pie/8m=pier^2Xh, then , pie/8 - pier^2=h, then, pie/8pie r^2=h, then, 1/8r^2=h, is this write?
then show this expressen as A=pier^2+pie/4r
area of base= pier^2
curved area side hX2pier=2pie r h
total area pier^2+2pie r h
pie r^2+2 pie r 1/8r^2
and finally A=pie r ^2+ pie/4r is this right?

then it says to find dA/r and d^2A/dr^2
which i have done 2pier-pie/4r^2 then 2pie+pie/2r^3
But finally i need to find the corresponding minimum values..but im not sure how too so this....
can you tell me how please?
Thank you
Apr 30th 2008, 09:58 AM
galactus

"Pie?".. So, it has to do with baking pies?.
Apr 30th 2008, 10:00 AM
Chez_

No lol the pie expression in maths , i dont know how to put it as a symbol on this
Apr 30th 2008, 10:13 AM
galactus

It's spelled 'Pi'. I had to, I'm sorry.

I assume when you say radius, the bin is cylindrical?.

I can tell you that the surface area is minimum when the height equals the diameter.

We have $V={\pi}r^{2}h=\frac{\pi}{8}$ ...[1]

The surface area, assuming the bin has a top and bottom, would be

$S=2{\pi}rh+2{\pi}r^{2}$ ...[2]

This is what must be minimized.

So, we solve [1] for h andf sub into [2]:

Solving [1] for h, we get $h=\frac{1}{8r^{2}}$

That goes into the surface formula:

$2{\pi}r(\frac{1}{8r^{2}})+2{\pi}r^{2}$

Differentiate:

$S'(r)=4{\pi}r-\frac{\pi}{4r^{2}}$

Set to 0 and solve for r and we get $r=\frac{2^{\frac{2}{3}}}{4}\approx{0.397}$

Plug that into h and it should be twice that.

$h=\frac{1}{8(.397)}=.794=\frac{2^{\frac{2}{3}}}{2}$

Sure enough. So, the surface area is minimum when the height equals the diameter.
Apr 30th 2008, 10:45 AM
Chez_

The bin only has a bottom,, does this change it?
Apr 30th 2008, 11:42 AM
janvdl

Quote:

Originally Posted by Chez_

The bin only has a bottom,, does this change it?

Exterior Area becomes:
$2 \pi r h + 1 \pi r^2$
Apr 30th 2008, 12:30 PM
janvdl

For only a bottom: (Using Galactus's post)

Quote:

Originally Posted by galactus

We have $V={\pi}r^{2}h=\frac{\pi}{8}$ ...[1]

The surface area, assuming the bin has only a bottom:

$S=2{\pi}rh+1{\pi}r^{2}$ ...[2]

This is what must be minimized.

So, we solve [1] for h and sub into [2]:

Solving [1] for h, we get $h=\frac{1}{8r^{2}}$

That goes into the surface formula:

$2{\pi}r(\frac{1}{8r^{2}})+1{\pi}r^{2}$

Differentiate:

$S'(r)=2{\pi}r - \frac{\pi}{4r^{2}}$

Set to 0 and solve for r and we get
$2 \pi r=\frac{\pi}{4 r^2}$
$r^3=\frac{1}{8}$
$r=\frac{1}{2}$

Plug that into h.

$h=\frac{1}{8(0.25)}=\frac{1}{2}$