# Thread: max volume of a box

1. So we have a box. Its width is , and its length is . I'm not given a height. The total surface area is sq. ft. I need the maximum volume. Easy enough if I find the formula for the volume, but I'm finding myself unable to do that. This isn't homework, it's just a problem I encountered and it's frustrating that I can't get it. Help anyone?

2.

3. 1) derive an equation for the surface area in terms of h and x.

2) derive an equation for the volume in terms of h and x.

3) these two equations will let you define x in terms of h.

4) obtain an equation for the volume in x ( or h)

5) differentiate the volume with respect to x (or h)

6) solve dv/dx=0 to obtain a value for x or dv/dh=0 to obtain a value of h

7) sub these values back into the volume formula and this should give you a max volume

4. Should've mentioned, it was meant to be done without calculus. And I know all those steps, but what I was doing didn't yield a function with any maximum, so it didn't work.

5. pfft why would we want to do it without calculus.

is that the exact replication of the question?

6. Originally Posted by organic god
pfft why would we want to do it without calculus.

is that the exact replication of the question?
Yeah, I know. I'd definitely use calculus personally.

It's from a pre-calc course, so I know they didn't intend for the student to use calculus. It was a homework problem I was helping someone with and unfortunately couldn't. I know full well how to do it, but whatever I was doing didn't yield a function with a maximum, it was just really screwy. It upsets me that I couldn't do it.

But yeah, all you're given is that the width is x, the length is 2.5x, and the total surface area is 54. They asked for a function to represent the volume, from which the max volume can be derived. As far as not using calc goes, its just a matter of throwing the correct function (once I have it) into a graphing calculator and having it find the maximum. So that is calculus...it's just that the calculator is doing it for the pre-calc student. Poor people without calculus...how do they survive?

7. best way i can think to do it, is when you have h in terms of x

the volume (2x^2)(h). will become a cubic.

completing the square on a quadratic is the best way but as we dont have that here. best thing to do is.

add some arbritary constant k to the cubic. doing this provides a vertical translation. at this point the cubic will touch the x axis and therefore a double root will exist for this function.

so you can write this as (x-a)(x-b)^2 = c(x^3)+d(x^2)+m(x)+g+K

then equate coefficents. long and tedious. just use calculus

8. Well what I was doing was... width is , length is , height is unknown, total surface area is . So... . So ?

Then according to that the volume is... V= ? But we definitely aren't getting a maximum out of this. Don't know where I'm going wrong.

9. The area is not quite right.

Area of the x by 2.5x faces = 5x^2.

Area of the x by h faces = 2xh

Area of the 2.5x by h faces = 5xh

10. ah...yeah knew it would be something stupid. thanks.

11. Originally Posted by Chemboy
ah...yeah knew it would be something stupid. thanks.
Happens to me every day, but then, I ofter miss the obvious.

12. Let’s look at the problem this way.

Let the dimensions of the box be a, b, and c. Then the AM–GM inequality gives

Now the surface area is and the volume is

Hence

i.e.

Hence the maximum volume is

13. I’m sorry, I made a mistake.

In applying AM–GM above, we would have if and only if As two of the sides of the box are and however, and it is impossible for the box to be a cube. Hence the max volume won’t be 27 (it will be less than that).

14. OK, let the dimensions of the box be

Then the surface area is

And the volume is

And we want to find the maximum value of without using calculus. Yes, it can be done.

15. Yeah, do it this way.

By AM–GM,

Hence the maximum volume is – which is attained when

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