Quote of the Day:
"Everything tries to be round."  -- Black Elk (Holy Man of the Oglala Sioux)

Objectives:
The student will learn to solve applied maximum minimum problems.

1. Intro
   We are going to spend a few days solving Applied 
     Maximum/Minimum Problems.
   
   I use this when I go to the Mall and try to decide where
     to park to minimize the distance to two stores.
   I used this when I parked at Daniel Morgan Middle School 
     to pick up my children (so that their distance to the car 
     would be a minimum).  
   
   Animals toward the poles have shorter limbs and 
     extremities and rounder bodies ...
     This minimizes surface area of the body (which 
     minimizes heat loss) – we're talking about penguins
     and polar bears.

   When blowing bubbles with various blowers, the bubbles 
     formed exhibit minimum surface area.

2. Experiment: 
   Items Needed: 
     Rectangular sheet of paper (8.5" x 11") for each 
      student
     Scissors
     Scotch Tape
     Rulers

   Give out a sheet of paper to each student.
   Have each student cut out squares from each corner 
     (these four squares must be the same size).
   Then fold up the four sides and tape to form a box.
   Have each student determine the volume of the box and 
     record results on the board or overhead or computer. 
   Place the results in two columns:

    Size of the Square cut out     Volume of box
    ---------------------------    --------------
           1" x 1"                  58.5 cu.in.
           1.5" x 1.5"              66.0 cu.in.
           3"x 3"                   37.5 cu.in.    
                 etc.

   Examine the data in the table to find the square that    
     would result in the maximum volume.  It should occur 
     between the 1.5" square and the 2" square. 

   Take time to look at various boxes and their volumes.  
   Have students hold up boxes with 1" square cut out, 3" 
     square, etc.

3. Now use calculus to solve the problem:

    Since we wish to maximize the volume of the box, we need to 
      determine a formula for the volume in terms of one variable.  
    Then we will take the derivative of V and set it equal to zero
      to find the relative maximum or minimums. 

    Let x = side of the square to be cut out.
    Then V = (11 – 2x) (8.5 – 2x) x
    Refer to the diagram.

4. Click here for Fifty Ways To Work A Problem

    
5. Assignment:
   Read section 6.2
   p. 348 (17, 19)

Click here for bubble blowing activity        

Click here for link to bubbles and minimal surfaces        

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