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

1. ## 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

2. "Pie?".. So, it has to do with baking pies?.

3. No lol the pie expression in maths , i dont know how to put it as a symbol on this

4. 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.

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

6. Originally Posted by Chez_
The bin only has a bottom,, does this change it?
Exterior Area becomes:
$2 \pi r h + 1 \pi r^2$

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

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.

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