Cylindrical Cost Problem?

Question

An open-top cylindrical can holds 16pi m^3 of water.

Material for the floor costs $2 per m^2.

Material for the side costs $1 per m.

Find the radius and the height of the most economical can.

Answer 1

Shouldn't the material for the side cost per m^2 (not per m)?

Answer 2

The cost of the cylinder is given by:

C = $2 * area of base + $1 * area of the side.

(I assume you meant that the cost of the material for the side is $1/m^2, not $1/m).

The area of the base is given by

A_base = pi*r^2, where r is  the radius of the base.

The area of the side is given by:

A_side = h * circumference = h * 2*pi*r

where h is the height.

Putting this together, we have that:

C (in dollars) = 2*pi*r^2 + 1*h*2*pi*r

Now, this equation gives the cost in terms of two variables, h and r, but we know that the cylinder has to hold 16*pi m^3 of water, so we know that the volume of the cylinder must equal this value.  The volume of a cylinder is just the area of the base times the height, so:

V = 16*pi m^3 = h*pi*r^2

16 m^3 = h*r^2

h = (16 m^3)/r^2

We can now use this relationship between h and r to eliminate one of these variables in the cost equation above:

C (in dollars) = 2*pi*r^2 + 1*(16/r^2)*2*pi*r, where r is measured in meters.

C = 2*pi*r^2 + 32*pi/r

Now take the derivative of this equation with respect to r, set the derivative equal to zero, and solve for the value(s) of r that give the extremal values (maximum and/or minimum) of the cost:

dC/dr = 4*pi*r - 32*pi/r^2

0 = 4*pi*r - 32*pi/r^2

divide through by pi, and multiply through by r^2 (which is ok, because r = 0 is not going to be a feasible value for r):

4*r^3 - 32 = 0

r^3 - 8 = 0

r^3 = 8

r = 2 meters.

Now go back and find the value of h corresopnding to this value of r:

h = (16 m^3)/r^2

h = (16 m^3)/( 4 m^2) = 4 m

The most economical can has a radius of 2 meters and a height of 4 meters.

You should check to make sure that this is actually the minimum value of the cost, and not a maximum.  The easiest way to check this is simply to plot the cost as a function or r, and confirm that there is a minimum, not a maximum at r = 2.

The way to check this without resorting to plotting the function is to determine the sign of the second derivative of the cost with respect to r at r = 2.  If the function is a minimum, then the second derivative will be positive, whereas if it is a maximum, the second derivative will be negative.  (It's zero if it's a saddle point.)

In this case,

C'' = 4*pi + 64*pi/r^3

At r = 2, C'' = 4*pi + 64*pi/8 = 12*pi, which is positive, so this is a minimum.