Optimization Problems

Question

A cylindrical can is to have a volume of 1L. 

a) Find the height and radius of the can that will minimize the surface area

b) What is the ratio of the height to the diameter? Do pop cans have a similar ratio? If not, why?

Answer 1

A cylindrical can is to have a volume of 1L.
a) Find the height and radius of the can that will minimize the surface area

 

1 liter = 1000 cm³

 

Let x = the radius of the top and bottom of the can

Let y = the height of the can

 

The volume of the can is:  V = (pi)x²y = 1000 cm³.  So y = 1000/(pi)x²

 

The surface area of the can is: S = 2(pi)x² + 2(pi)xy

 

Substituting 1000/(pi)x² for y, the surface area is:

 

S = 2(pi)x² + 2(pi)x*1000/(pi)x² = 2(pi)x² + 2000x^-1

 

Take the derivative of S wrt x:

 

S' = dS/dx = 4(pi)x - 2000x^-2

 

S' = 0 = 4(pi)x - 2000x^-2

 

2000x^-2 = 4(pi)x

 

2000/4(pi) = x³

 

x = (2000/4(pi))^1/3 = 10(1/2(pi))^1/3 ≈ 5.42 cm

 

y = 1000/(pi)x² = 10.84 cm

 

Check:

 

V = (pi)(5.42)²(10.84) = 1000

 

b) What is the ratio of the height to the diameter?

 

The ratio of the height (y) to the diameter (2x) = 1

 

Do pop cans have a similar ratio? If not, why?  You're on your own for this question!  ;-)

Answer 2

This is a cylinder so we know that the base and top are A = π*r² where r is the radius of the cylinder. The side area is the length of the outer edge of the base times the height (which we will call h) or A = 2πr*h. So the total area is

 

A = 2π*r² + 2πr*h = 2π (r² + rh).

 

The volume is just:

 

V = πhr² = 1

 

So, using the volume we have h = 1/(πr²). We can substitute this into Area to simplify our expression.

 

A = 2π (r² + 1/(πr)).

 

Now we take the derivative of A with respect to r, so that we can find critical points:

 

A' = 2π ( 2r - 1 / (πr²) ).

 

Set it equal to zero to find critical points, discarding the constant 2π since it is irrelevant:

 

0 = 2r - 1 / (πr²),

 

2r = 1 / (πr²),

 

r³ = 1/(2π),

 

r = 1 / (2π)^1/3.

 

So our diameter is

 

D = 2πr = 2π * 1 / (2π)^1/3 = (2π)^2/3.

 

Now put this in the volume equation to find h.

 

πh(1/ (2π)^1/3)² = 1,

 

πh/ (2π)^2/3 = 1,

 

h= 2^2/3/π^1/3.

 

Now we can find the ratio:

 

Ratio = h/D = (2^2/3/π^1/3 ) / (2π)^2/3 = ( 1/π^1/3 ) / π^2/3 = 1 / π