# Homework Help: Optimizing surface area of a silo

1. Oct 19, 2008

### elitespart

A silo has cylindrical wall, a flat circular floor, and a hemispherical top. If the cost of construction per square foot is twice as great for the hemispherical top as for the walls and the floor, find the ratio of the total height to the diameter of the base that minimizes the total cost of construction.

There was another problem involving the silo which said: for a given volume, find the ratio of the total height to the diameter of the base that minimizes the total surface.

For this I wrote the equation for volume and solved for h and then I plugged that into the equation for area of the silo, derived, set equal to zero and found that radius and height both equal cubed route of 3v/5pi which would make the ratio: h+r/2r = 1.

What do I have to do differently for this question? Thanks.

2. Oct 19, 2008

### elitespart

any advice on this? I'm really stuck and this hmwk is due tomorrow.

3. Oct 19, 2008

### Dick

The cost is proportional to twice the area of the top plus the area of the other parts. That's the only thing you have to change. Minimize cost instead of area.

4. Oct 19, 2008

### elitespart

So I'm just minimizing A = (pi)R^2 + 2(pi)RH + 4(pi)r^2?

5. Oct 19, 2008

### Dick

Yes.

6. Oct 19, 2008

### elitespart

Thx brotha

7. Oct 19, 2008

### elitespart

So I got R and H = (3V/11pi)^(1/3). Would the ratio be 2H + R / 2R or 3/2?

8. Oct 20, 2008

### HallsofIvy

"So I got R and H = (3V/11pi)^(1/3)"

What does that mean? You got what for R? The "ratio of height to diameter of base" is H/(2R).

9. Oct 20, 2008

### elitespart

well it says total height so the height of the cylinder + the hemisphere which would be
H + R/(2R). I thought that since the hemisphere costs twice as much it would be 2H. And I got R = H = (3V/11pi)^(1/3). So where'd I mess up? Thanks.