# A metal cylindrical can is to have a volume of 3.456pi cubic feet. How do you find the radius and height of the can which uses the smallest amount of metal?

##### 1 Answer

The smallest area occurs when we have a radius of

#### Explanation:

Let us set up the following variables:

# {(r, "Radius (feet"), (y, "Height of can (feet)"), (A, "Surface Area of can (sq feet)") :} #

We want to vary the radius

Then the volume is fixed:

# pir^2h = 3.456pi #

# :. r^2h = 3.456 #

# :. h = 3.456/r^2 #

And, The Surface Area is given by:

# A=2pirh + 2pir^2 #

# :. A=2pir(3.456/r^2) + 2pir^2 #

# :. A=(6.912pi)/r + 2pir^2 #

Differentiating wrt

# :. (dA)/(dr)=(6.912pi)(-1/r^2) + 4pir #

# :. (dA)/(dr)=(-6.912pi)/r^2 + 4pir #

At a critical point,

# :. (-6.912pi)/r^2 + 4pir = 0 #

# :. -6.912pi + 4pir^3 = 0 #

# :. r^3 = 1.728 #

# :. r = 1.2 #

With

# A=(6.912pi)/1.2 + 2pi(1.2)^2 #

# :. A=5.76pi + 2.88pi #

# :. A=8.64pi ~~ 27.143#

We should check that this value leads to a minimum (rather than a maximum) area. As the size of the can is finite this should really be intuitive. We could calculate the second derivative and verify that

graph{(6.912pi)/x + 2pix^2 [-2, 15, -500, 500]}

Hopefully you can visually confirm that a minimum does indeed occur when