2003 2 unit trial, last qu. (1 Viewer)

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Hey people!
As usual i'm having difficulty with the later stages of teh paper. Here's teh last qu. I'll be grateful for any help, you don't necessarily have to answer it.

Qu. 10 b)

A farmer is building a wheat silo in teh shape of a closed cylinder of radius r metres and height h metres. The silo is to be made from galvanised iron sheeting and is to have a capacity of 300 m cubed.

i) Find an expression for the height of the silo in terms of r

ii) Show that the surface area A, of teh silo is given by the equation:

A= 2pi r^3 + 600
-------------------------
r

iii) Hence, find the min. area of galvanised iron sheeting to make the silo, leave answer in exact form

 

[Nelson.]

i)n.r^2.h=300
h=300/n.r^2

ii)Surface area = 2.n.r^2 + 2n.r.h

=2.n.r^2 + 2.n.r(300/n.r^2)
=2.n.r^2 + 600/r
=(2.n.r^3 + 600)/r

too lazy to type out iii) just differeniate it then solve for r from da/dr=0...then test for the nature(max/min)

 

[Xcelz]

All my dots = x

i) Volume of cylinder= pi.r^2.h
pi.r^2.h = 300
h= 300/pi.r^2

ii) A of Cylinder= 2.pi.r(r+h)
= 2.pi.r^2 + 2.pi.r.h
= 2.pi.r^2 + 2.pi.r(300/pi.r^2)
=2.pi.r^2 + 600/r (Cross multiply)
=(2.pi.r^3 + 600)/r

iii) A = (2.pi.r^3 + 600)/r
= 2.pi.r^2 + 600r^-1 (Split into two fractions cause I'm to lazy for the quotient rule and I divided everything by 'r')

dA/dr = 4.pi.r - 600r^-2

To confirm as a minimum.

d^2A/dr^2 = 4.pi + 1200r^-3
= 4.pi + 1200/r^3
(Since r > 0, second differential > 0 = minimum)

For stat point dA/dr = 0
4.pi.r - 600r^-2 = 0
4.pi.r = 600/r^2
4.pi.r^3= 600
r^3 = 600/4.pi

r = Cube root of (600/4.pi)
r= 3.63 approx

Sub r back into A formula.

A = (2.pi.(3.63) + 600)/3.63
= 171.57 m^2

Might have left a couple of things and it may be wrong but hope it helps.