Problem of finding the volume of a box.

In summary, the conversation discusses finding the maximum volume of a box that can be accepted by airlines, given specific dimensions. The attempt at a solution involves using partial derivatives, but the problem is meant to be solved without their aid. Eventually, the correct solution is found by assuming that the length and width of the box are equal and using this to simplify the volume equation. The maximum volume is found to be (31 - h/2)^2 times h.
  • #1
naaa00
91
0

Homework Statement



The airlines accept a box if length + width + height <= 62. If h is fixed, show that the maximum volume (62-w-h)wh is V = h(31 - h/2)^2.

The Attempt at a Solution



Well, I multipied:

V(h) = (62-w-h)wh
V(h) = 62wh - w^2h - wh^2

dv/dt = 62w - w^2 - 2w = 60w - w^2

0 = 60w - w^2
w = 60.

But this can not be. This is wrong. I tried other things but all of my attemps are wrong...
 
Physics news on Phys.org
  • #2
What you have done makes no sense at all. You write "V(h)" when V depends on both w andh as variables. And then you write "dv/dt" when there is NO "t" variable at all!

A differentiable function of two variables will have max or min only where its gradient is 0 or, equivalently, [itex]\partial V/\partial w= 0[/itex] and [itex]\partial V/\partial h= 0[/itex].
 
  • #3
Thank you for your answer, HallsofIvy.

"dv/dt" well, yes. That was a typo from my part. My bad. Clearly volume does not depent on t...

I see that you have written partial derivative notation. I am assuming that your suggestion is to use partials, since volume seems to depend on two variables, namely h and w. But this exercise is in a problem set of a single-variable calculus textbook. I guess this can be solved that way, but for the moment I think the textbook expects me to solve the problem without the aid of partials. (I don't know know anything of partials. I only know something about its notation.)

My thought is that since V = HLW, I would get something of the form V = h times M^2, if I assume L = W = M^2, and then substitute. But I don't know how to prove this assumption and I have the feeling that it is wrong...
 
Last edited:
  • #4
But the problem statement says h is fixed. I am not sure what does the textbook means with "fixed" (I am not an english native speaker.) I suppose it means that h is not a variable, since it is "fixed"; it does not change. If that is the case, then V depends on L and W. But the problem ask me to show that (62-w-h)wh is V = h(31 - h/2)^2. Here "V" seems to be in function of h; a some sort of contradiction. I am a little bit confused...
 
  • #5
Well, certainly I was confused yesterday. Happily now I undestand the problem:

V = (62 - h - w)wh,

V = 64wh - h^2w - w^2h.

dv/dw = 62h - h^2 - 2wh,

0 = 62h - h^2 - 2wh,

w = 62h - h^2/ 2h

w = 31 - h/2.

It turns out that L = W, because L = 62 - h - w, or L = 62 - h - (31 - h/2), or L = 31 - h/2.

Then, if we let W^2 = LW, it is clear that V = W^2H. Finally we have (31 - h/2)^2 times h.

Thanks! See you next time!
 

Related to Problem of finding the volume of a box.

1) What is the formula for finding the volume of a box?

The formula for finding the volume of a box is length x width x height.

2) How do I measure the length, width, and height of a box?

You can measure the length, width, and height of a box using a ruler or measuring tape. Make sure to measure in the same unit (e.g. inches, centimeters) for accuracy.

3) Can I use the same formula for irregularly shaped boxes?

No, the formula for finding the volume of a box only works for regular rectangular boxes. For irregularly shaped boxes, you may need to use a different formula or break the shape down into smaller regular boxes and add their volumes together.

4) Do I include the thickness of the box's walls in the volume calculation?

No, the volume of a box only includes the space inside the box, not the thickness of the walls. If you need to calculate the volume of the entire box including the walls, you will need to add the volume of the walls separately.

5) Can I find the volume of a box with only two dimensions given?

No, you need all three dimensions (length, width, and height) to calculate the volume of a box. If you are missing one dimension, you will not be able to accurately find the volume.

Similar threads

  • Calculus and Beyond Homework Help
Replies
33
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Calculus and Beyond Homework Help
Replies
20
Views
2K
  • Introductory Physics Homework Help
Replies
25
Views
499
Replies
7
Views
2K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
13
Views
27K
Back
Top