Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

**allseeingskink****Member**- Registered: 2005-11-02
- Posts: 4

G'day everyone,

This question is from year 12 (I'm australian).

I've got exams this week and this problem has come up and I have noooo idea. Its from a previous year's exam paper.

Question:

Some person is constructing a time capsule, it will be a right circular cylinder of hieght 'h' and radius 'r', with hemispherical caps of radius 'r'

a) Express V in terms of r and h

my answer (p = pi):

V = (4/3)pr^3 + pr^2

b) The total volume of the capsule will be 8000cm cubed.

i. Show that h = 8000/(pr^2) - 4r/3

My answer was just to sub V = 8000 into my above formula and rearrange in terms of 'h'.

ii. The values which r may take lie in an interval. Find the end points of this interval, correct to two decimal places.

My answer (I dont know if this is correct!)

the function of h in terms of r above gives an asymptote at r = 0, and intersects r-axis at r= 12.41, therefore interval is:

(0, 12.41)

If that right??

Moving on, part c.

c) The material for the cylindrical part of the capsule costs 2 cents per sq. cm. The material for the hemispherical caps costs 3 cents per sq. cm. Find an expression for C cents, the total cost of the materials for the capsule, in terms of r.

I do:

C = 12pr^2 + 4prh

sub h = .... from previous question

C = 12pr^2 + 4pr(8000/(pr^2) - 4p/3)

C = 32000/r + 20r^2 /3

Is that right?

Anyway, now it asks to draw a graph of C against r for an appropriate domain. Label any asymptotes, you are not required to show the co-ordinates of any turning point.

My graph is like a hyperbola with no turning point until well outside the domain.

Now, the next question, it asks to use calculus to find the value of r, correct to two decimal places, for which C is a minimum.

But my equation has a minimum at 69.3 or so. Which is well out of the domain that I gave above (0, 12.41). That domain might be wrong? Its what I figured out myself. There must be a mistake earlier on.

Thanks to anyone who can help me out.

*Last edited by allseeingskink (2005-11-02 19:34:26)*

Offline

**allseeingskink****Member**- Registered: 2005-11-02
- Posts: 4

ARGH! I'm such a complete idiot. After staring blindly at the question for a looong time, I found what the problem is.

In the equation for Cost in terms of 'r', its:

C = 12pr^2 + 32000/r - (16/3)r^2

I stupidly tried to simplify the 12pr^2 and the (16/3)r^2 and completely ignored pi when I did. So it ended up C = 32000/r - (20/3)r^2 when it should have been:

C = 32000/r - (12p - 16/3)r^2

This makes the graph of C against r look more like a skewed parabola, which of course has a turning point which can be found using calculus, rather than the hyperbola I was getting.

Argh, sorry, I feel like such an idiot. I'd looked at the question for an hour before posting this, and only noticed just then.

Exams are stressing me out too much I think. :s

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,615

So, you helped yourself out then, right? Taka a bow!

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

Offline

**mikau****Member**- Registered: 2005-08-22
- Posts: 1,504

I go through this same episode every day. lol.

A logarithm is just a misspelled algorithm.

Offline