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?

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?

Tutors, please sign in to answer this question.

Olney, MD

1 liter = 1000 cm^{3}

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)x^{2}y = 1000 cm^{3}. So y = 1000/(pi)x^{2}

The surface area of the can is: S = 2(pi)x^{2} + 2(pi)xy

Substituting 1000/(pi)x^{2} for y, the surface area is:

S = 2(pi)x^{2} + 2(pi)x*1000/(pi)x^{2} = 2(pi)x^{2} + 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) = x^{3}

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

y = 1000/(pi)x^{2} = 10.84 cm

Check:

V = (pi)(5.42)^{2}(10.84) = 1000

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

Castle Rock, WA

This is a cylinder so we know that the base and top are A = π*r^{2} 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*r^{2} + 2πr*h = 2π (r^{2} + rh).

The volume is just:

V = πhr^{2} = 1

So, using the volume we have h = 1/(πr^{2}). We can substitute this into Area to simplify our expression.

A = 2π (r^{2} + 1/(πr)).

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

A' = 2π ( 2r - 1 / (πr^{2}) ).

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

0 = 2r - 1 / (πr^{2}),

2r = 1 / (πr^{2}),

r^{3} = 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

Steven M.

Premium Test Prep and Subject Tutor - New York City UWS

New York, NY

5.0
(257 ratings)

Michael E.

H.S. Mathematics Tutor (Certified NJ Teacher)

Saddle Brook, NJ

5.0
(131 ratings)

John P.

Math, Science, Technology, and Test Prep Tutor

Fort Lee, NJ

5.0
(688 ratings)

- Math 9396
- Calculus 1 474
- Integration 109
- Precalculus 1481
- Derivatives 203
- Differentiation 122
- Calculus 2 334
- Multivariable Calculus 25
- Calculus 3 181
- Vector Calculus 34

## Comments