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[chrisso23]

I was just wondering if someone would be able to help me with this question which i found the other day (sorry for the length):

"A farmer wishes to build a silo. The shape of the silo must be cylindrical with a conical "cap". The height of the cone is to be equal to the radius of it's base. The cone and base of the cylinder are to be made of a metal costing 20% more, per square metre, than the metal used for the cylinder walls.
The volume of the cylindrical part of the silo must be 54 cubic metres. For technical reasons the height of the cylinder must lie between 2 and 10 metres. Find the dimensions which give the least cost.
Note: The area of the curved surface of a cone of radius x and height y units is Pi*x*square root of (x squared + y squared)."

Thanks very much in advance

 

[wogboy]

First draw a diagram of the whole thing so you know what's going on before you get stuck into the maths.

if we let s1 be the surface area of the base, s2 the surface area of the curved cylinder wall, and s3 the surface area of the cone at the top, r to be the radius of the base (and cone height), and h to be the cylinder height:

s1 = pi*r^2
s2 = 2*pi*r*h
s3 = pi*r*sqrt(r^2 + r^2) = pi*sqrt(2)*r^2

the cost c, of the silo is given by:

c = 1.2*(s1 + s3) + s2

(since the base and cone material cost 1.2 times that of the cylinder walls)

c = 1.2(pi*r^2 + pi*sqrt(2)*r^2) + 2*pi*r*h
= 1.2*pi*(1 + sqrt(2))*r^2 + 2*pi*r*h ........ (A)

but we know that the volume of the cylinder equals 54 m^3, so

pi*r^2*h = 54.
r^2 = 54/(pi*h)
r = sqrt(54/(pi*h))
= sqrt(54/pi) * h^(-1/2) ...(B)

(ignore the negative root of r since r is positive)

if we eliminate r from equations (A) and (B), we will get:

c = 1.2*pi*(1 + sqrt(2))*(54/pi)*h^(-1) + 2*pi*h*sqrt(54/pi)*h^(-1/2)

so

c = 1.2*pi*(1 + sqrt(2))*(54/pi)*h^(-1) + 2*pi*sqrt(54/pi)*h^(1/2)

therefore,

dc/dh = -1.2*pi*(1 + sqrt(2))*(54/pi)*h^(-2) + pi*sqrt(54/pi)*h^(-1/2)

if we set dc/dh = 0, we get:

1.2*pi*(1 + sqrt(2))*(54/pi)*h^(-2) = pi*sqrt(54/pi)*h^(-1/2)

bringing the h's on one side:

h^(-1/2)*h^(2) = 1.2*pi*(1 + sqrt(2))*(54/pi)/(pi*sqrt(54/pi))

therefore,

h^(3/2) = 64.8*(1 + sqrt(2))/sqrt(54*pi)

(finally cleaned up that horrible mess on the RHS )

so,

h = [64.8*(1 + sqrt(2))/sqrt(54*pi))]^(2/3)
= 5.24 m

this is within the acceptable range of 2<=h<=10, so we can leave h at that value.

To find r, just simply use equation (B) above, and sub h=5.24 into it,

r = sqrt(54/pi)/sqrt(h)
= 1.81 m

 

[:: ck ::]

i got dizy from reading ur post but i think its rite