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#1 2007-08-14 14:01:51

Mel
Guest

Optimisation

A rectangular box is to be made by the following requirments
- The length must be one and a half times the width
- The twelve edges must have a total length of 6m
Find the dimensions of the box that maximises the volume

:S

I know the answers are 30 000, 14 000

#2 2007-08-14 17:25:35

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Optimisation

the first thing is to write out what we know in equation form, we have a box, with length L, height H and width W. We are told that

L = 1/2 W and we also are told the sum of the twelve edges is 6m (i assume that means, 6 meters)

there are 4 of each type of length, so we have

4W + 4L + 4H = 6;

we know that

Volume = L*W*H, now we use the two equations above to eliminate to variables so that only one remains. We have that L = 1/2W so  Volume = 1/2W*W*H, moreover we have  4W + 4L + 4H = 6 so (replacing L with 1/2 W) 4W + 2W + 4H = 6, so 6W + 4H = 6 so H = 3/2(1 - W), replacing this for H in the equation Volume = 1/2W*W*H we get Volume = 3/4W*W*(1-W) so volume = 3/4W^2 - 3/4W^3.  So now we want to find the highest point on the graph of y = 3/4 x^2 - 3/4 x^3 where y is the volume and x is the width. At this peak point, the graph will flatten out and turn back down, thus the peak point occurs somewhere where the slope of the graph, the derivative, is equal to zero.

We have that y = 3/4 x^2 - 3/4 x^3, differentiatng we get dy/dx = 3/2x - 9/4x^2 . If this is equal to zero then 3/2x - 9/4x^2 = 0 so 3/2*x*(1 - 3/2x) = 0; this is true if either x = 0 or if x = 2/3. So at x = 0 and x = 2/3 the slope is zero, but which is at the top of the peak? (the highest point on the graph)

if either of these values are at a peak point of the graph, the graph will be concaved downward there, that is, the second derivative (3/2 -9/2x) will be negative. At x = 0 it is positive, at x = 2/3 it is negative, so 2/3 is a maximal point so W = 2/3

note now that we can use the equations L = 1/2 W and 4W + 4H + 4L = 6 to solve for the remaining variables.


A logarithm is just a misspelled algorithm.

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#3 2007-08-14 18:15:03

Mel
Guest

Re: Optimisation

=S

woah.. ok then but i dont get how the answer is 30 000, 14 000

#4 2007-08-14 18:16:29

Mel
Guest

Re: Optimisation

hahaha never mind i was looking at the wrong answer
thanks

#5 2007-08-14 21:33:47

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: Optimisation

Mikau's method is good, but his numbers are wrong. One of the initial equations should be L = 3/2W.

Following through with that change gets you that W = 2/5.


Why did the vector cross the road?
It wanted to be normal.

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#6 2007-08-15 07:44:48

mikau
Member
Registered: 2005-08-22
Posts: 1,504

Re: Optimisation

x_x my bad! I thought it said one half the width.


A logarithm is just a misspelled algorithm.

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#7 2007-08-16 00:29:09

George,Y
Member
Registered: 2006-03-12
Posts: 1,379

Re: Optimisation

length, width
l=1.5w
then how about the hight? Okay, set it as h.

then we know that 4(l+w+h)=6

so 2.5w+h=1.5

Now we wanna maximize V=1.5w w h    just under the condition of 2.5w+h=1.5.

The simplist way to do this is to gimmikly apply an inequality

We know that for a, b and c

and that

and that

the equality stands in previous two when a=b=c.

"abc" in the the second clause is more or like volume formula.

let's see 1.5w w h ....... a=1.5w, b=w, c=h--------wait! a and b can never be equal in this way, meaning we cannot find the very maximum point in this way.

However, there is another way, the smart way. Watch this:

a= b=1.25w and c=h.

Then
V= (1.5/1.25²) 1.25²w²h≤  (1.5/1.25²) ((2.5w+h)/3)^3

the equality stands when 1.25w=1.25w ( definately) =h

The maximum volume is then 1.5/12.5=0.12 when w=0.4 and h=0.5, and of cause, l=0.6:D

Last edited by George,Y (2007-08-16 00:33:46)


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