# Math Help - Optimization help :)

1. ## Optimization help :)

hey all, newbie here.

I have got a problem regarding optimization in 3D:

"A silo is to made from cylinder surmounted by a hemisphere. The cost of construction per square foot of surface area is twice as great for the hemisphere as for the cylinder. Determine the dimensions to be used if the volume is fixed and the cost of the construction to be minimum. Neglect the thickness of the silo and waste in the construction. The floor of the silo is not included in this construction"

I easily can solve the 2dimentional version of this problem, but having little problems here in 3D. Especially on the part of the price, can I assume the the price for the hemisphere to be 2 units and for the cylinder 1 units, so I can proceed the algebra?

thanks a lot,
Tuugii

2. Originally Posted by Tuugii
...

"A silo is to made from cylinder surmounted by a hemisphere. The cost of construction per square foot of surface area is twice as great for the hemisphere as for the cylinder. Determine the dimensions to be used if the volume is fixed and the cost of the construction to be minimum. Neglect the thickness of the silo and waste in the construction. The floor of the silo is not included in this construction"

...
Hello,

the volume of the silo is:

$V=\pi r^2 \cdot h + \frac23 \pi r^3$ . Rearrange to get $h=\frac{V}{\pi r^2} - \frac23 r$

The surface area is:

$s=2\pi r \cdot h + 2 \pi r^2$

The price is

$p=2\pi r \cdot h + 2 \cdot 2 \pi r^2$ . Now plug in the term for h and you'll get a function of the price depending on r:

$p(r)=2\pi r \cdot \left( \frac{V}{\pi r^2} - \frac23 r \right) + 4 \pi r^2$ . After few steps of rearranging you'll get:

$p(r) = \frac{2V}{r} + \frac83 \pi r^2$

You'll get an extreme (minimum or maximum) value of the price if p'(r) = 0:

$p'(r) = -\frac{2V}{r^2} + \frac{16}3 \pi r$ . Now solve p'(r) = 0 for r. I've got: $r=\sqrt[3]{\frac{3V}{8 \pi}}$

3. ## thanks:)

many thanks for the post!

that is the same answer I got. You had used the same method that I used, where we both assumed the price would be as 1 unit on one and 2units for the other.

I was just wondering why can we use this? 1:2 is the ratio of the prices, but we have an equation expressing the value of the price which is: P(r).

Can you please explain me this?

Tuugii

4. Originally Posted by Tuugii
...
I was just wondering why can we use this? 1:2 is the ratio of the prices, but we have an equation expressing the value of the price which is: P(r).
...
Tuugii
the second statement tells you that. the ratio is the relationship of the prices of each part of the object while your P(r) is the price of your whole object. Ü

5. Originally Posted by kalagota
the second statement tells you that. the ratio is the relationship of the prices of each part of the object while your P(r) is the price of your whole object. Ü

exactly, here is the thing I am wondering about:

P(r) - is the value of the price
1:2 - is the ratio of the price.

But we had expressed the Value n terms of the ratio:

P(r)=k(1:2)

1:2 is the ratio, is can be 1$and 2$ or 500$and 1000$, we don't know; wouldn't it influence the value of the Price P(r) directly?

thanks a lot,
Tuugii