Assuming that the cost of 16-gauge steel sheet metal is $25.00 per square meter, what is the ideal radius and height of a cylindrical drum in meters if you want to minimize the cost of producing the oil drum?

Answer: Radius = 32.7 cm and height = 65.5 cm

Explanation:

An oil drum must have a volume of 218 liters or 218*1000cm^3 = 218,000 cm^3

The price is per square meter, so if we reduce the surface of the oil drum, we will pay less:

So we want to play with the measures of the oil drum in such a way that the surface is minimized.

Now, first, the volume of the oil drum (a cylinder) is:

V = pi*r^2*h

where pi = 3.14, r is the radius and h is the height.

and the surface is:

S = 2*pi*r^2 + 2*pi*r*h

we know that:

pi*r^2*h = 218,000 cm^3

r^2*h = 218,000cm^3/3.14 = 70,096.5 cm^3

now we can write

h = 70,096.5 cm^3/r^2

now we can replace it in the surface equation:

S = 2*pi*r^2 + 2*pi*r*h

= 6.28 * (r^2 + 70,096.5 cm^3/r)

So we want to minimize this, we can derivate it and find the zero:

S' = 6.28 (2*r - 70,096.5 cm^3/r^2) = 0

2r = 70,096.5 cm^3/r^2

r^3 = (70,096.5 cm^3) / 2

r = ∛ ((70,096.5 cm^3) / 2) = 32.7cm

And then the height is:

h = 70,096.5 cm^3/r^2 = 70,096.5 cm^3 / (32.7cm) ^2 = 65.5cm