Expert: Paul Klarreich - 12/8/2007

Question
http://en.allexperts.com/q/Calculus-2063/Maximum-minimum-problems-2.htm?zIr=5
For that problem, I believe there may be a mistake.  I am confused as to how you have h= (108-x^2)/x instead of (108-x^2)/(4x)

Answer
Questioner:   Casey
Category:  Calculus
Private:  No
 
Subject:  Previously Posted Question
Question:  http://en.allexperts.com/q/Calculus-2063/Maximum-minimum-problems-2.htm?zIr=5
For that problem, I believe there may be a mistake.  I am confused as to how you have h= (108-x^2)/x instead of (108-x^2)/(4x)
=========================================
Hi, Casey,

If you think I blew it, there is no need to worry about hurting my feelings.  You can go ahead and say so.  
 
In this case, the problem should go:

Dimensions are l,w,h

l = w, so volume = wwh

base = w^2.
side = wh, four of them

A = w^2 + 4wh = 108

Solve for h:

4wh = 108 - w^2

   108 - w^2
h = ---------  << Yes, I had just 'w' in the denominator.  Definitely an error.
      4w

v = w^2h

       108 - w^2
V = w^2 ------------
          4w


       108 - w^2
v = w ------------
          4


    108w - w^3
V = ------------
        4

       108 - 3w^2
dV/dw = ----------
           4

Set that = 0 (top only)

108 = 3w^2

w^2 = 36

w = 6, and the rest goes as before.  It appears the error did not affect the result.

But thank you for pointing this out.  It is always gratifying to find that people actually read my stuff, even when they find errors.  I will try to post a correction.