# problem on finding max value (URGENT)

• May 28th 2009, 04:31 AM
matref
problem on finding max value (URGENT)
I was wondering what would be the way to answer this question. I know that trial error is one way but i also know that there is a differential calculus like method. But i am not sure about how to do it. Could some one pleaes help me!!!!!

"Find the length of the cuboid which has a maximum permissible length of 1.070m and also has the maximum possible volume"

NOTE: must ensure that sum of the maximum length and the corresponding circumference (circumference of plane perpendicular to length)

All help is appreciated.
• May 28th 2009, 12:15 PM
HallsofIvy
Quote:

Originally Posted by matref
I was wondering what would be the way to answer this question. I know that trial error is one way but i also know that there is a differential calculus like method. But i am not sure about how to do it. Could some one pleaes help me!!!!!

"Find the length of the cuboid which has a maximum permissible length of 1.070m and also has the maximum possible volume"

NOTE: must ensure that sum of the maximum length and the corresponding circumference (circumference of plane perpendicular to length)

All help is appreciated.

So you have a rectangular solid with one side of length x and two of length y and must have x+ 4y= 1.070? Or sides of length x, y, z with x+ 2y+ 2z= 1.070?
The volume is \$\displaystyle xy^2\$ in the first case, xyz in the second.

In the first case, x= 1.070- 4y so \$\displaystyle xy^2= 1.070y^2- 4y^3\$. Differentiating that, \$\displaystyle 2.140y- 12y^2= y(2.140- 12y)= 0\$ which gives either y= 0 or y= 2.140/12= .1783. y= 0 gives a package of volume 0- obviously the maximum. So y= .1783 and then x= 1.070- 4(.1783)= 1.070- 0.7132= 0.3568. That makes the maximum volume \$\displaystyle (.3568)(.1783)^2= 0.011342989552.\$

I don't think there is enough data to do the other case so I hope this was what you meant! Are you mailing a package?