Expert: Paul Klarreich - 8/23/2009

Question
the problem is as follows:

I need to design a can for a sports drink that will minimize the amount of material needed to save on cost. the can needs to hold 50 cubic inches of the drink (27.2 fluid ounces). What should the height and radius of the can be to minimize the surface area?

Answer
Questioner: Jake
Country: United States
Category: Calculus
Private: No
Subject: calculus question
Question: the problem is as follows:

I need to design a can for a sports drink that will minimize the amount of material needed to save on cost. the can needs to hold 50 cubic inches of the drink (27.2 fluid ounces). What should the height and radius of the can be to minimize the surface area?
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Hi, Jake,

Is this your first attempt at Max-min problems?  If so, the scheme is something like this:

1. Identify the variable(s) in the problem -- the things that can be changed.  Give them names.

2. Write the thing to be Max-ed(or min-ed) in terms of the variable(s).

3. If there are more than one, determine a relationship between the variables (called a 'CONSTRAINT') that will eliminate all but one.  Use a diagram, use your life experience, use your general knowledge and brilliance, do whatever you have to.  This is probably the hardest part.

4. Now differentiate, set the derivative = 0, and solve.  Be sure to check for endpoint max-mins.

AND, Please check the archives for other M-M examples.  There are a lot of them.  Click BROWSE PAST ANSWERS.
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You make such a can from a rectangle wrapped around and sealed:

Area = circumference * height

, plus two semicircles.

Area of each = pi r^2

Variables:

r = radius of base of can
h = height of can
A = surface area.

Relation:

A = 2 pi rh + 2 pi r^2

Constraint:

V = 50 = pi r^2 h

Solve for h in that:
    50
h = -----  
   pi r^2

Substitute:

A = 2 pi r(50/pi r^2) + 2 pi r^2

A = 100/r + 2 pi r^2

A = 100 r^-1 + 2 pi r^2

You can handle the rest.