The Mathematics Education Department
EMT 725, Lisa F. Garey
Click here for the Problem Statement
This is THE classic calculus lidless box problem. The question? Maximize the volume. Now that I've been led to discover the A.M.-G.M. inequality, I don't need calculus to solve this problem.
Multiplying by 4/4 we have 
By the A.M.-G.M. inequality, 
with equality iff 4x = 25-2x, so x = 25/6 = 4.17
Thus the volume the box can never be larger than 1157.4 cu. units and this occurs when x= 4.17 units.
Another investigation is using Algebra Expresser to graph the curve and then trace the maximum point.
Algebra Expresser can also help find what shape boxes could be created to hold specific volumes.
| Volume | x1 | x2 |
| 200 cu. units | 0.4 | 10.35 |
| 400 cu. units | 0.75 | 9.25 |
| 800 cu. units | 1.75 | 7.3 |
Generalizing the result to an nxn square sheet of cardboard, we have
with equality iff 4x = n-2x, so x=n/6.
If the cardboard is 20 by 25 units, we get volume = x(20-2x)(25-2x). When we use the A.M.-G.M. inequality, the equality condition becomes problematic because we end up with 4x=20-2x=25-2x.
Return to Lisa's EMT 725 Problems