Expert: Paul Klarreich - 10/18/2007

Question
A manufacturer wants to design an open-top box having a square base and a surface area of 108 square inches.

What dimensions will provide a box with maximum volume?

Length of base : inches

Width of base : inches

Height of box : inches

Give all of your answers correct to two decimal places.

Answer
Questioner:   Ashley White
Category:  Calculus
Private:  No
 
Subject:  Optimization << I changed it to conform to the description of other problems of this type.  This web site has an archive of problems and you can find a lot of examples of this type.  Look around -- you'll find them.
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Question:  A manufacturer wants to design an open-top box having a square base and a surface area of 108 square inches.

What dimensions will provide a box with maximum volume?

Length of base : inches

Width of base : inches

Height of box : inches

Give all of your answers correct to two decimal places
.......................................
Hi, Ashley,

In these 'max-min' problems, you want to:

1. Determine what the variables are and give them names.
2. Determine the quantity to be maximized (minimized?) and express it in terms of the variables. [this might take some thought and some life experience, etc.]
3. If the quantity is a function of more than one variable, use the secondary conditions (constraints) to eliminate all but one of the variables.
4. Differentiate, set the derivative equal to zero, solve, and then test your set of 'candidate' points for max or min.
.......................
Variables:

L = length of base.
W = width .......
H = height.

V = volume, TO BE MAXIMIZED.

V = LWH.

But there are constraints:

1. the base is a square, so  L = W.  << get rid of L
2. the surface area is 108 sq in.

Area = base + four sides.

108  = W^2 + 4(WH)

108  = W^2 + 4WH
   108 - W^2
H = ---------    << CORRECTION TO PREVIOUS ANSWER.
       4W

               108 - W^2
V = LWH = W W (-----------)
                  4W
               
V = W(108 - W^2)/4

   108W - W^3
v = ----------
       4

Ready to go.

       108 - 3W^2
dV/dW = ---------
           4


Set that equal to zero.

108 - 3W^2 = 0

108 = 3W^2

36 = W^2

W = 6   <<  W = -6 has no meaning here.

That will be it, basically.  You will now go back to the constraints to find the other dimensions:

L = 6

   108 - 36
H = --------- = 12
       6