Add Maths Project for form 4 2006 .. Help seriously needed ... Time is running out !!?

Question

I was jus wondering if u noe how to do this question ..

The Muhibbah Company is a manufacturer of Cylindrical aluminiun tins . The manager plans to rdues the cost of production . The production cost is propotional to the are of the aluminium sheet used . The volume that each tin can holds is 1000cm cubed { 1 litre } .

1 ) Detrenima the value of h { height } , r
{radius } and hence calculate the ratio of h/r aka height over radius , when the total surface area of each tin is minimum .

2 ) The top and bottom pieces of the tin of height ,h cm are cut from the square-shaped aluminium sheets . Determine the value of r , h , and hence calculate the ratio of h over r so that the total are of the aluminium sheet used for making the tin is minimum .

I really need this urgently .. do you know the answer ? and the method of calculation ? I need 2 methods of answering for each questions . Many thanks in advance .. { dateline is on thursday { 9 nov 06 }

Answer 1

Total surface area of a tin can = A = 2pr^2 + 2prh
Volume of a tin can = V = pr^2h = 1000 cm^3
Therefore, h = 1000 / pr^2

Substituting this into the area equation gives :
A = 2pr^2 + 2pr * 1000 / pr^2
= 2pr^2 + 2000 / r

Now take the derivative of A with respect to r :
dA / dr = 4pr - 2000 / r^2

Set this equal to zero to find the minimum :
4pr - 2000 / r^2 = 0
This can be rearranged to show that : 1000 / pr^2 = 2r, which equals h, as shown above.
Continuing to find r, we get : r = (500 / p) ^ (1/3), or approx. 5.42 cm.

Now h = 2r = 2 * (500 / p) ^ (1/3), or approx. 10.84 cm.

Thus, ratio is h / r = 2r / r = 2

I'm a bit confused about the second question, but I think what you want is a square sheet of aluminium, such that all 3 pieces of a can will fit, with the minimum of wastage. This would be a square sheet with the dimensions : (perimeter of can) by (height + 2 * radius).

Thus, Perimeter = Height + 2 * Radius, because the sides of a square are equal.
Therefore, 2*pi*r = h + 2r,
so, h = 2r(pi - 1)

But, V = pi*r^2*h = 1000, so h = 1000 / (pi*r^2)

Equating the 2 values for h gives :
2r(pi - 1) = 1000 / (pi*r^2)
from which we get : r = {500 / [pi(pi - 1)]} ^ (1/3)

So now we know that h = 2r(pi - 1), so that after substituting for r, we get :
h = 2 * (pi - 1) ^ (2/3) * (500 / pi) ^ (1/3)
And we find that the ratio h / r = 2r(pi - 1) / r = 2(pi - 1)

Well, that was a bit of speculation, but even if I've got it wrong, then perhaps you can adapt it to your needs.

Edit
Can't help feeling I've misunderstood the second question the first time around, but on re-reading it many times, I still don't understand it.

Answer 2

Don't bother doing the homework at all.
You lack integrity and clearly can not handle responsibility. You happily pass the buck and are keen to take credit for the work of others.
Mathematics is not a prerequisite subject for the welfare queue. Perhaps you could swap the class for Self-Victimization - 101 and get a real head start...

Answer 3

those are the steps I take to freshen up a chain of questions: a million) study heavily the textual content fabrics e book 2) study heavily my type notes 3) answer all questions properly 4) I now have not were given any project answering those self same questions even as they take position on the subsequent attempt.

Answer 4

What Jane said..........