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Call me stupid, but someone could please solve this for me????
A piece of wire 100cm long is going to be cut into several pieces and used tp construct the skeleton of a rectangular box with a square base.
(a) what are the dimensions of the box with the largest volume?
(b) what are the dimensions of the box with the largest Surface Area?
-Thanks.
MHENK
OOOOOOOOO, OOOOOOOO
This is a calculus question, I took Calculus....
Let me think....
OK a box has 12 edges, so 100 cm divided by 12 is 8 and 1/3 cm per side. The surface area is then 416 and 2/3 cm^2, and the volume is also 416 and 2/3 cm^2
...don't know for sure if thats right, but it should be.
mlewis -- theoretically that would work, but I am trying to maximize the volume of the box. I am not saying that all edges of the box are the same, they can change. Tahnks anyway though
I think you need to take the derivative of the volume of the rectangle, using the value of 100 cm. I don't have my book with me, and its been a while since I did one of those Max/Min problems, but I think you do something like that....
The volume of a rectangle should be length*width*height...
Width and hight are the same so...
V=L*W*H...
V=L*W^2...
Input 100 cm as V...
100=L*W^2
THEN I'm lost...I forget if you take the derivative now or after you solve for L??
If I recall correctly, if all the edges add up to 100cm, then it will always have the same volume and surface area...at least, it seems like it should!!!!
Therefore my answer is correct...
So...
V=L * W ^ 2
The derivative of V=(L*2W)+(0*W^2)...
Which is V=L2W
L=V/(2W)
Input L back in the function
V = (V/(2W)) * (W^2)
Input 100 cm as V
100=(100/2W)) * (W^2)
I'm very confused...
You need to solve for L and W, Input 100 cm somewhere, and at some point take the derivative. I can't help you any more than that. Sorry:(
what's this do for you?
(4l+4w+4h)=100
l+w+h=25
v=l*w*h
SA = 2lh+2lw+2wh
Mlewis-Your answer is correct if and only if the box is a cube. I need to deal with any other type of box. The length and width are the same, but the height varies. This gets
l^2+h=25
v=h*l^2
SA=4lh+2l^2
now to maximine surface area and volume.... what do i do?
You have to take the derivative/integral of Volume/SA. Can I check my Calculus book when I get home, and tell you tomorrow??
Its been almost a year since I did maximizing probs.
That's not really a general VB question... Isn't it?
I thought General VB Questions stood for General, Very Bad, Questions. Sorry, i'll keep this stuff in the right place.....
Seriously though, no one reads the maths forums consistantly, so when I have a question I need solved quickly, I post it here and let the VB Supergeniuses work on it...
-MHENK
It would be if someone suggests MHENK writes a VB proggie that calculates and graphs all possible volume and surface areas as the bottom square changes from 0cm square to 100cm square.
Does this mean that the length of all the edges must add up to 100cm, or does it mean that the maximum length of an edge is 100cm?Quote:
Originally posted by MHENK
A piece of wire 100cm long is going to be cut into several pieces and used tp construct the skeleton of a rectangular box with a square base.
(a) what are the dimensions of the box with the largest volume?
(b) what are the dimensions of the box with the largest Surface Area?
a rectangular box with a square base is a cubeQuote:
Originally posted by MHENK
Call me stupid, but someone could please solve this for me????
A piece of wire 100cm long is going to be cut into several pieces and used tp construct the skeleton of a rectangular box with a square base.
(a) what are the dimensions of the box with the largest volume?
(b) what are the dimensions of the box with the largest Surface Area?
-Thanks.
MHENK
100 = 12 / x
x = 100 / 12
x = 8 1/3
The volume of a cube:
Length ^ 3
the surface area
6 * Length ^ 2
So the volume would be:
578.703703703703703703703703703704
and the surface area would be
416.666666666666666666666666666667
That's the only maximum value because a rectangle with a square base is, well a cube ;)
oh no, sorry my bad
it's a cuboid
i get it now ;)
The following is not perfect but gives you an idea.
It gives me the following result:
Max Volume: 578.703426
8.32999999999987x8.32999999999987x8.34000000000027
Max Area: 416.6666
8.32999999999987x8.32999999999987x8.34000000000027
you can change the accuracy by decreasing the increment in the following lines:
Width = Width + 0.01
Depth = Depth + 0.01
But it will never give you a perfect result.
Code:Dim Width As Double
Dim Height As Double
Dim Depth As Double
Dim Volume As Double
Dim Area As Double
Dim MVWidth As Double
Dim MVHeight As Double
Dim MVDepth As Double
Dim MaxVolume As Double
Dim MVArea As Double
Dim MAWidth As Double
Dim MAHeight As Double
Dim MADepth As Double
Dim MAVolume As Double
Dim MaxArea As Double
Width = 0
Depth = 0
Do Until Height < 0
'Calculate the height
Height = (100 - (((Width * 2) + (Depth * 2)) * 2)) / 4
'Calculate Volume
Volume = Width * Depth * Height
'Calculate Area
Area = ((Height * Depth) * 4) + ((Width * Depth) * 2)
Debug.Print "volume: " & Volume, "Area: " & Area
If Volume > MaxVolume Then
MVWidth = Width
MVHeight = Height
MVDepth = Depth
MaxVolume = Volume
MVArea = Area
End If
If Area > MaxArea Then
MAWidth = Width
MAHeight = Height
MADepth = Depth
MAVolume = Volume
MaxArea = Area
End If
Width = Width + 0.01
Depth = Depth + 0.01
Loop
Debug.Print "Answer:"
Debug.Print "Max Volume: " & MaxVolume
Debug.Print MVWidth & "x" & MVDepth & "x" & MVHeight
Debug.Print "Max Area: " & MaxArea
Debug.Print MAWidth & "x" & MADepth & "x" & MAHeight
Cutting wire into 12 equal length pieces and making a cube is the correct answer. The proof has been posted in the maths forum.
Making a cube seems like a good guess for maximizing the volume. Without working on the problem, it is not obvious (to me anyway) that the same shape will maximize both area and volume, no matter what that shape is.
you find the extreeme values for the function at the solution for the derivative(function) =0:Quote:
v=h*l^2
SA=4lh+2l^2
now to maximine surface area and volume.... what do i do?
v=hl^2
l^2+h=25
l^2=25-h
v=(25-h)h
v=25h-h^2
Dv=25-h=0 => h=25
V=whl
4l+4w+4h=100
l+w+h=25
three unknowns two equations, subproblem:
V=xwh where x is constant, gives the an expression for max one of the variables using extreemevalues at Derivate=0
w+h=25-x
V=xw(25-x-w)=25xw-x^2w-xw^2
DV=25x-x^2-2xw=0 => w=(25-x)/2 ; w=0 or w=25 would give a flat box, the extreemvalue is maximi
going the same way with h we get h=(25-x)/2 => w=h, and for l: l=(25-x)/2 assuming x was say substitute for w. => w=h=l=25/3
A=2wh+2wl+2hl
l+w+h=25
three unknowns two equations, subproblem:
A=2wh+2wx+2hx where x is constant, blahblah see above
A=2(25-x-h)h+2(25-x-h)x+2hx
DA=2(25-x-h)-2x+2x=>h=50-2x; h=0 or h=25 would give a flat box..blahblahblah(See above)... =>w=h=l=25/3
duh, i'm late as usual :( Well just a different solution here :p
what an earth is BORK?
I second the motion!! What is BORK!?!?
Oh and da_silvy, I am really dying to know the phrase encrypted in your signature, but my cracking skills are a bit rusty... if you must keep it a secret, just pm/email me :p
ehehheheh wouldn't you like to know.
;)
I am thinking....
This stuff is more complicated.
:)
:confused:Quote:
Originally posted by haravinth
I am thinking....
This stuff is more complicated.
:)
what stuff?Quote:
Originally posted by haravinth
I am thinking....
This stuff is more complicated.
:)
He probably read Ked's post.
It takes a while to realise that Ked has a pseudo maths generator which spits out lines of totally random mathematical looking gibberish just to confuse people.
Quote:
Originally posted by Arbiter
It takes a while to realise that Ked has a pseudo maths generator which spits out lines of totally random mathematical looking gibberish just to confuse people.
Really? I gotta get me one for VB ;)
OK after coding myself a random "crap" generator, I have determined that is is the same one Microsoft uses to make their software! This is the Windows XP source code:
VB Code:
Note the terrible programming concepts and the use of the eternally cursed GoTo!!!!
:D
yes, that's good code
Never ending loop.
Never ending goto recursion :eek:
Try it -- it WILL end...eventually!
In fact, it will only run once!
Read through carefully and see why...
in fact,
it won't even run
sub or function not defined :rolleyes: :(
DUH!!!! :rolleyes: :rolleyes: :rolleyes:
Of COURSE they're not defined! I'm just talking about the loop
meh, got no idea :rolleyes:
*looks at the thread's subject*
:D
What's THAT supposed to mean??!Quote:
Originally posted by Jotaf98
*looks at the thread's subject*
:D
Are you trying to say we're off subject or something?!? :p
Kedaman: Just noticed your post. It seems to be correct. At least you got the right answer.
However, iusing solving simultaneous equations seems a bit complicated and unnecessary. See the Thread in the Maths Forum.
If you start with one unknown variable (say X), which is the decimal fraction of the wire used for base and top of the parallelopiped, you can get a fairly direct solution with just a bit of algebraic donkey work.
My little "endless loop" above is not endless for a very crazy reason: it will eventually do a BSOD, so the system must reboot, and thus the loop is ended! :p
yep, it is a bit complicated.
No, it's just that the subject says "a real quick question"... it doesn't seem to be that quick to me :pQuote:
Originally posted by mlewis
What's THAT supposed to mean??!
Are you trying to say we're off subject or something?!? :p
it's been solved like 4 times, he just hasn't noticed :eek:
Lol you didn't really think I'd read trough an entire page of replies to a maths question did you? :D