Expert: Paul Klarreich - 5/29/2008

Question
You're designing an open top card board box for a purveyor of nuts. The top will be made of clear plastic, but the plastic box top designer is handling that. The box must have a square base and 2 cardboard pieces that divide the box into 4 sections for the almonds, cashews, pecans, and walnuts. Given that you want a box with a volume of 72 cubic inches What dimensions will  minimize the total cardboard area and thus minimize the cost of the cardboard? Whats teh total area of the cardboard?
This the the problem I got in class but i don't know where to start or how to do it.

Answer
Questioner:   Nicole
Category:  Calculus
Private:  No
 
Subject:  Min and Max
Question:  You're designing an open top card board box for a purveyor of nuts. The top will be made of clear plastic, but the plastic box top designer is handling that. The box must have a square base and 2 cardboard pieces that divide the box into 4 sections for the almonds, cashews, pecans, and walnuts. Given that you want a box with a volume of 72 cubic inches What dimensions will  minimize the total cardboard area and thus minimize the cost of the cardboard? Whats teh total area of the cardboard?
This the the problem I got in class but i don't know where to start or how to do it.
...................................................
Hi, Nicole,

Start by making a diagram -- a good one!  Here is a cross-section of the base:

+---------+---------+
|         |         |  
|         |         |  
|         |         |  
|         |         |  
+---------+---------+
|         |         |  
|         |         |  
|         |         |  
|         |         |  
+---------+---------+


Then decide what the variables are; give them names.  Say something about each. IN WORDS.

Let h = height of box.
   s = one side of base. (Both are the same -- it is a square.)
   A = total cardboard area. TO BE MINIMIZED.    

Express the variable to minimize:

There will be a base and six 'walls' to make:
The base is s^2
Each wall is hs.
The total is
A = s^2 + 6hs

You need a constraint. [You always do -- otherwise you would make the box so tiny it costs almost nothing.]

V = 72.

But V = s^2h

So s^2h = 72 and we can solve for h:

h = 72/s^2

Plug back in:

A = s^2 + 6hs

A = s^2 + 6(72/s^2)s

A = s^2 + 432/s

Now you are basically done -- differentiate, set = 0, solve, etc. just like any other max-min exercise.  Good luck.