# Need Help on max&min problems

1. Nov 4, 2004

### oreon

Hi Everybody

I have two max and min word questions,I didnt understand well how to solve this type question,thats why, I would like you to help me and show me how to solve these questions.I know some begining steps and know all solution way.But the problem is I dont know how to use or do those steps.Could you show me but not only show also teach me with explaining the steps sothat I can see what you did.I will have a small test tomorrow, and there will be afew min max questions.I want a learn how to those and after you show me, I will work on similar questions like them.
I will wait your responde and helps...

Here are two questions.
----------------------

1)My first question is:
A silo has a hemisphrerical roof,is cylindrical sided and has a circular floor, all made of steel. find the dimensions of a silo with a volume of 755 cubic meters that uses the least amount of steel to build. justify your answer with calculus. so it is asking minimize surface area.

note:there is a picture of a silo but there isnt anything on the picture except there is r on the top and on the bottom.also there is a h left side of silo.

My approach is=
So the two parameters are the height, h, of the silo, and
the radius, r, of the silo. The volume will be the volume
of the cylinder plus the volume of the hemisphere:
V = (2/3)pi r^3 + pi r^2 h
So V is given in the problem, allowing me to solve
for h in terms of r.

The total area is the sum of the area of the hemisphere and
the area of the cylindrical part:

A = 2pi r^2 + 2pi rh
Substitute for h in the area equation, then take the
derivative to find dA/dr. Set it to zero and solve for r.
Then back-substitute to find h.

2)Second question is:

8. Nov 5, 2004

### Galileo

V = (2/3)pi r^3 + pi r^2 h
So, solving for h:
$$h=\frac{V-(2/3)\pi r^3}{\pi r^2}$$

plug this in the area (without the floor, do the correct one with the floor area yourself):
$$A = 2\pi r^2 + 2\pi rh=2\pi r^2 + 2\pi r\frac{V-(2/3)\pi r^3}{\pi r^2}$$
rewrite it with a bit of algebra:
$$A=\frac{2}{3}\pi r^2+\frac{2V}{r}$$
taking the derivative and setting it equal to zero:
$$A'=\frac{\frac{4}{3} \pi r^3-2V}{r^2}=0$$
gives $r=(\frac{6V}{4\pi})^{1/3}$
Notice that A'<0 if r is smaller than that critical value and A'(r)>0 if r is greater than that critical value. So you have an absolute minimum.
Since V=755 m^3, r=7.12 m.
then plug r back into:
$$h=\frac{V-(2/3)\pi r^3}{\pi r^2}=$$
to find h.
Just like you said in your first post.

(Surprise: h=0.)

The fence problem is correct as far as I can tell.
For completeness though, you should check and add that x=18 really gives a minimum C, not merely a point where the derivative is zero.

Last edited: Nov 5, 2004
9. Nov 5, 2004

### oreon

Thank you very much galileo ,you are a great person.

Last edited: Nov 5, 2004