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Differential Calculus, Area, Volume, Box

Oliver

New member
Joined
Dec 10, 2016
Messages
9
Hello everyone

I'm struggling a little with this one, any help would be greatly appreciated, thank you.

Question:
A chemical manufacturer has commissioned a series of stainless steel open top rectangular boxes with square ends, what is the least area of stainless steel required to give a volume of 4m^3

a) Using differential calculus calculate the area of stainless steel required

b) Using the second derivative determine if this is a minimum volume
 

HallsofIvy

Elite Member
Joined
Jan 27, 2012
Messages
4,716
If you have been "struggling with it for some time" certainly should be able to show us what you have done! That would give us an idea of what hints and suggestions you need. This is an "open top rectangular box with square ends". Do you understand what that means? A "rectangular box" means that it has 6 sides but since this is "open top" only 5 of them are made of steel. I do find "square ends" to be ambiguous. What, exactly, is an "end"?

I am going to assume that the ends are the top and bottom and take them to be squares of side length x. Taking the sides to have height y, sides are rectangles with width x and height y.

The four sides, then, have total area 4xy and the bottom has area \(\displaystyle x^2\). Since the box is "open" we do not count the area of the top. The total area is \(\displaystyle 4xy+ x^2\).

We are told that the volume is "4 cubic meters". The volume of a rectangular solid is "width times length times height" so \(\displaystyle x^2y= 4\). The problem asked you to minimize \(\displaystyle 4xy+ x^2\) subject to the constraint \(\displaystyle x^2y= 4\). There are couple of different ways to do that. Since you have not shown what you have tried, I do not know which would be appropriate for you.

(I "assume that the ends are the top and bottom" because if the sides are squares then the top and bottom must be also. In that case the "rectangle" is a cube and there is only one cube with volume 4. There would be no need to "minimize" anything.)
 

Oliver

New member
Joined
Dec 10, 2016
Messages
9
I said I'd been struggling with it, you added the "for some time" bit yourself :)

I didn't show what I'd done because I'm not even sure if I'am starting correctly, I might be thinking about this completely wrong and wanted a fresh perspective without my input.

I didn't write/create the question, just copied it out word for word, the way I read it I imagined it shaped like this:


rectangular box with square ends, the green bit being open.

I thought it would be something like:

Surface Area=2X[SUP]2[/SUP]+3XY
Volume=X[SUP]2[/SUP]Y = 4
So Y=4/X[SUP]2[/SUP]
Therefore, Surface Area=2X[SUP]2[/SUP]+3X*(4/X[SUP]2[/SUP]) which then simplifies to = 2X[SUP]2[/SUP]+(12/X)

I think I then differentiate it which would be 4X - (12/X[SUP]2[/SUP]) and this is pretty much where I got to. I haven't had a question like this before so I have no prior knowledge of something even a little like this to help me work through it, I'm just guessing here.

Sorry I haven't wrote that with the proper equation formatting like you have, I don't know how to do that on here and its 3:30am so I'll figure that out tomorrow. I know there are easier ways of getting the answer to the question but I have to use differential calculus and show the working out involved.
 

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Otis

Full Member
Joined
Apr 22, 2015
Messages
957
Surface Area = 2X[SUP]2[/SUP]+3XY

Volume=X[SUP]2[/SUP]Y = 4

Y=4/X[SUP]2[/SUP]

Surface Area = 2X[SUP]2 [/SUP]+ 12/X

differentiate it ... 4X - 12/X[SUP]2

[/SUP]this is pretty much where I got to.
Your work matches mine, to this point.

I have no prior knowledge of something even a little like this to help me work through it ...
As you have not seen a minimization problem worked before, here is the part that you're missing.

A minimum (or maximum) function value occurs at a value of X which causes the derivative to equal zero.

In other words, to obtain the answer, you find the Real value of X which causes the derivative to become zero, and then you substitute it for X in the surface-area function.

In part (b), I do not understand why they ask about a minimum volume because the volume is constant, in this exercise; there are no other volumes.
 

Otis

Full Member
Joined
Apr 22, 2015
Messages
957
function

Google keywords video function minimum derivative, to find worked examples of function minimization/maximization.

I just did; those hits from the khan academy will likely be decent. :)

By the way, you don't need fancy formatting, to type basic expressions in your posts:

Area = 3xy + 2x^2

y = 4/x^2

Derivative = 4x - 12/x^2
 

Oliver

New member
Joined
Dec 10, 2016
Messages
9
Thank you Otis, that is exactly what I was missing.


Surface Area = 2X[SUP]2[/SUP]+3XY
Volume=X[SUP]2[/SUP]Y = 4
Y=4/X[SUP]2[/SUP]
Surface Area = 2X[SUP]2[/SUP] + 12/X
Differentiate it = 4X - 12/X[SUP]2[/SUP]


When 4X - 12/X[SUP]2[/SUP] = 0 then X = [SUP]3[/SUP]√ 3 or 1.442249570307408

So put this back into the SA = 2([SUP]3[/SUP]√ 3)[SUP]2[/SUP] + 12/[SUP]3[/SUP]√ 3 = 12.4805m[SUP]2[/SUP]


Surface Area = 12.4805m[SUP]2
[/SUP]
b) From what I read I think this is something to do with the second derivative being positive or negative number.
So first derivative = 4X - 12/X[SUP]2[/SUP]
second derivative = 24/X[SUP]3 [/SUP]+ 4
So 24/([SUP]3[/SUP]√ 3)[SUP]3[/SUP]+ 4 =12.00001
This is positive therefore giving a minimum value. So the minimum surface area of the box is 12.4805m[SUP]2
[/SUP]
Agree? Disagree? :)
 

ksdhart2

Full Member
Joined
Mar 25, 2016
Messages
929
Everything looks fine to me, except a minor error in the derivative. If you copied that value from a calculator, it's highly likely that's just a "floating point error." The value of the second derivative at the point x=cbrt(3) is exactly 12. In any case, that very small inaccuracy doesn't change the fact that the second derivative is positive, and therefore your analysis is correct. Good job.
 

Oliver

New member
Joined
Dec 10, 2016
Messages
9
Thank you ksdhart, well spotted with the slight error It was from using an online calculator... proves you properly checked it haha ;)

I really appreciate the help from every one as I wasn't sure I was even on the right track, thank you.
 
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