Expert: Paul Klarreich - 8/7/2009

Question
I'm not sure how to set up the following word problem. I know that once I've got an appropriate equation, I want to take the partial derivative to arrive at the answer, but I can't figure out how to set it up. Thanks for your help, here is the prob:

The US Postal service carries small packages, but when the box gets big, the post office measures the length x, width y, and height z, if x + 2y + 2z > 100, then the box is rejected. What measurements x, y, z give rise to the acceptable box with the largest volume?

Answer
Questioner: Drew
Country: United States
Category: Calculus
Private: No
Subject: volume maximization
Question: I'm not sure how to set up the following word problem. I know that once I've got an appropriate equation, I want to take the partial derivative to arrive at the answer, but I can't figure out how to set it up. Thanks for your help, here is the prob:

The US Postal service carries small packages, but when the box gets big, the post office measures the length x, width y, and height z, if x + 2y + 2z > 100, then the box is rejected. What measurements x, y, z give rise to the acceptable box with the largest volume?
......................................................
Assuming

V = xyz, and

x = 100 - 2y - 2z

then
V = (100 - 2y - 2z)yz

V = 100yz - 2y^2z - 2yz^2

W = V/2 = 50yz - y^2z - yz^2   << get rid of a useless 2.

Now  
Wz = 50y - y^2 - 2yz   << Wz is the partial w.r.t. z.

Wy = 50z - 2yz - z^2

Now we would like both of those to be zero:

50y - y^2 - 2yz = 0
50z - 2yz - z^2 = 0

50 - y - 2z = 0
50 - 2y - z = 0

50 = y + 2z
50 = 2y + z
-------------- Subtsract:  (You are right -- I can't spell)
0 = y - z

So y = z, so make:

50 - y - 2y = 0
50 - 3y = 0

y = z = 50/3

x = 100 - 2(50/3)

x = 100 - 100/3

x = 200/3

There are your dimensions:

Length = 200/3, width = height = 50/3