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frazhm
19th Dec 2007, 08:34
Here is a problem set for my son in his GCSE maths class (math - for our american friends!!)

A shopkeeper asks a company to make some trays.
A net of a tray is made from a piece of card measuring 18cm x 18cm

The Shopkeeper says
"When the area of the base is the same as the area of the four sides, the volume of the tray will be a maximum"

Investigate.

Now in his answer so far he has logically worked out various volumes and a maximum volume appears around a base of about 11cm square. But I feel that we could devise a quadratic equation.

Along the lines of height of tray =x
length = 18-2x
width = 18-2x
area of base = (18-2x) squared


and so on

So guys and gals what do you think????
When I try to work it out and then use the values in the old quadratic formula I end up with a daft result. I feel I must be missing something


Many thanks

Frazhm and Happy Christmas

Hoping
19th Dec 2007, 09:03
ok, my x and y might be different to yours, so I'll define them first. x is the edge length of the base of the tray, y is the height of the tray.
Because you're limited to a net of size 18cm, y is determined exactly by your choice of x, i.e. x+2y = 18, meaning y = (18-x)/2.
An expression for the volume V of the tray is easy to find. It is the area of the base x^2, multiplied by the height of the tray, y. We already have an expression for y in terms of x (y = (18-x)/2) and so the volume is given by
V = yx^2 which is (18x^2 - x^3)/2
To find a maximum of V, you could evaluate V for different values of x, ranging from 1cm to 17cm, and test which value gives maximum V - I guess that is as far as he would be expected to go for GCSE maths.
The 'proper' way to do it is to differentiate V with respect to x:
dV/dx = (36x - 3x^2)/2
At a maximum of V, dV/dx = 0. This is standard calculus.
An obvious value for x for which dV/dx = 0 is x=0, but that is unrealistic, so you can dismiss it. Another value is x = 12, which you can work out by rearranging (36x - 3x^2)/2 = 0 for x:
multiply both sides by 2 to get
(36x - 3x^2) = 0
add 3x^2 to both sides to get
36x = 3x^2
divide both sides by x to get
36 = 3x
divide both sides by 3 to get
12 = x
The answer to your problem is x = 12.

PS Don't give him the answer outright, use what you now know to try and guide him in the right direction. Plus, if they haven't been taught differentiation yet, don't let him submit the differentiation based solution as the teachers will know somebody else solved the problem for him...

ScottyDoo
19th Dec 2007, 09:13
I'm confused - are you trying to build a tray or a net???

Track Coastal
19th Dec 2007, 09:16
When the area of the base is the same as the area of the four sides

Whoops. I read that as 1 side (ie a cube). Edited due to my inability to RTFQ (read the f***ing question).

D SQDRN 97th IOTC
19th Dec 2007, 10:30
Track Coastal
So if X=6, what is the area of the base, and what is the area of the four sides?
Are they equal?
No. Base is 6*6. Four sides area is 4(6*6).
So volume of box is 6*6*6 = 36*6 = 216
If you take Hoping's maths, then volume of box is 12*12*3 = 124*3 = 372.
Area of base = 144. Area of 4 sides is 4(12*3) = 4 * 36 = 144.
Funny that.

Basic differential calculus says that when the differential of a function = 0, then that function is either at a local maximum or minimum. Hoping has discarded the 0 answer, as this is the minimum value of the function.
Second differential gives inflection points in the function etc......
I would hope Track Coastal that you are not a teacher of any maths !
:)

ZH875
19th Dec 2007, 10:34
Track Coastal
So if X=6, what is the area of the base, and what is the area of the four sides?
Are they equal?
No. Base is 6*6. Four sides area is 4(6*6).
So volume of box is 6*6*6 = 36*6 = 216
If you take Hoping's maths, then volume of box is 12*12*3 = 124*3 = 372.
Area of base = 144. Area of 4 sides is 4(12*3) = 4 * 36 = 144.
Funny that.

Basic differential calculus says that when the differential of a function = 0, then that function is either at a local maximum or minimum. Hoping has discarded the 0 answer, as this is the minimum value of the function.
Second differential gives inflection points in the function etc......
I would hope Track Coastal that you are not a teacher of any maths !
:)

D SQDRN 97th IOTC, I would hope that you are not a teacher of any maths !, as I estimate that 12*12*3=144*3=432

D SQDRN 97th IOTC
19th Dec 2007, 10:38
fair cop

it must be the effects of the booze from last night

12*12 is 144 of course.
(But I did get it right when I said the area of the base was 144 later on !)

Now, where did I put those new batteries for my old casio fx-82?

BlooMoo
19th Dec 2007, 13:46
When I try to work it out and then use the values in the old quadratic formula I end up with a daft result. I feel I must be missing something

The quadratic equation you need is given by Hoping

dV/dx = (36x - 3x^2)/2, where dV/dx = 0

this gives the quadratic equation

(3/2).x^2 - 18.x = 0

To solve it use the quadratic formula x= (-b+-sqrt(b^2-4ac))2a where from above

a= 3/2
b=-18
c=0

plug that it and you'll get x = 6 +- 6

i.e x = 0 or 12 as per Mr Hoping and therefore y=3

Keeping in mind the question, the shopkeeper has said that

When the area of the base is the same as the area of the four sides, the volume of the tray will be a maximum

We can verify using the same choice of variables :

Area of base = x^2

Area of each side = xy

therefore the shopkeepers statement is equivalent to asserting the volume is maximised when

x^2 = 4xy

From the above solutioni

12^2 = 4.12.3

i.e 144 = 144

so the optimal solution of x=12, y=3 satisfies x^2=4xy and so the shopkeeper is correct.

frazhm
19th Dec 2007, 14:29
Guys I am truly humbled - I just happen to mention my son's maths problem and within six hours I have seemingly accessed superior mathematical brains from all over the world!!!

How cool is that!!!

Many thanks to all
my son will be pleased

Merry Christmas and a Happy New Year

Frazhm

gearpins
19th Dec 2007, 15:28
1.area of cardboard= 18x18=324^cm
2.area of sum of sides=area of base---!
3.ht of tray=x
4.length of base=y
5.y+2x=18
6.sum of the area of 4 sides=4(x.y)
7.which is =area of base=y^2
8.y^2=4x.y in other words y=4x
9.substitute value of y in the 4th step: ~ 4x+2x=18
10.6x=18 ~x=18/6=3cm hence y=12cm

Forkandles
19th Dec 2007, 15:33
1.area of cardboard= 18x18=324^cm
2.area of sum of sides=area of base---!
3.ht of tray=x
4.length of base=y
5.y+2x=18
6.sum of the area of 4 sides=4(x.y)
7.which is =area of base=y^2
8.y^2=4x.y in other words y=4x
9.substitute value of y in the 4th step: ~ 4x+2x=18
10.6x=18 ~x=18/6=3cm hence y=9cm

must try harder, gearpins ;)

gearpins
19th Dec 2007, 15:41
thank you my friend:\ I suffer from fast finger freddy syndrome or something like that...so with the corrected value of y we have
area of sides=4(3.12)~4x36=144
area of base= 12x12=144.:)

D SQDRN 97th IOTC
19th Dec 2007, 15:44
gearpins

but the question was not to find out the dimensions of the box when the area of the sides equals the area of the base; it was to prove that when this condition was satisfied, the volume of the box was at its maximum.

gearpins
19th Dec 2007, 15:50
my understanding is that when the 2 areas are equal,it will produce a tray of max vol. as a given.
correct me if I am wrong

Mallan
19th Dec 2007, 15:53
How about smomething a bit less brain damaging.



This can tell you your age and how many times a week you would like to eat out



Wow.. this is amazing... it got my age right and the number of times I would
like to eat out per week and it worked !!!



Some of you older ones may need to get the calculator out...............I
did











YOUR AGE BY EATING OUT






Don't tell me your age; you'd probably would tell a lie anyway - but
your waiter may know!


YOUR AGE BY DINNER & RESTAURANT MATHS






DON'T CHEAT BY SCROLLING DOWN FIRST!
It takes less than a minute. Work this out as you read ...
Be sure you don't read the bottom until you've worked it out!
This is not one of those waste of time things, it's fun.(ish)


1. First of all, pick the number of times a week that you would like to
go outto eat.
(more than once but less than 10)


2. Multiply this number by 2 (just to be bold)


3. Add 5


4. Multiply it by 50


5. If you have already had your birthday this year add 1757...
If you haven't, add 1756.


6. Now subtract the four digit year that you were born.


You should have a three digit number


The first digit of this was your original number. (I.e., How many times
youwant to go out to restaurants in a week.)


The next two numbers are


YOUR AGE ! -- (Oh YES, it is!!!)


THIS IS THE ONLY YEAR (2007) ITWILL EVER WORK, SO SPREAD IT AROUND WHILE
IT LASTS

forget
19th Dec 2007, 16:10
I don't know what 'net of a tray' means. Does this happen only in Manchester? :confused:.

gearpins
19th Dec 2007, 16:35
frazhm,
ckeck your p.m.:)

BOFH
19th Dec 2007, 16:41
Mallan
The first digit of this was your original number. (I.e., How many times
youwant to go out to restaurants in a week.)

The next two numbers are
YOUR AGE ! -- (Oh YES, it is!!!)

Don't try this on people who are over 99.

BOFH

UniFoxOs
19th Dec 2007, 16:56
It's my birthday today - do I add 1756 or 1757??

UFO

PS:-
THIS IS THE ONLY YEAR (2007) ITWILL EVER WORK, SO SPREAD IT AROUND WHILE IT LASTS

For the non-mathematicians, it will work any year, you just need to change the 1756/1757 constants.

BlooMoo
19th Dec 2007, 17:02
my understanding is that when the 2 areas are equal,it will produce a tray of max vol. as a given.

Fair point actually, but in the context of an exam question on mathematical subjects the above is not 'given'.

It is normal for any 'given' to be made explicit by the paper's author - eg "In answering you may assume that ....etc, etc,".

In this case the examiner is introducing a 3rd person who is making a statement which that 3rd person is asserting *not* the examiner. The examiner is asking the student to 'investigate' i.e. treat the shopkeeper's statement as a hypothesis.

gearpins
19th Dec 2007, 18:45
I suppose you are right, when you put it that way.I was thinking as a kid would be expected to think...:8

Forkandles
19th Dec 2007, 18:48
If I was running a sheetmetal shop and this 'shopkeeper' didn't supply me with at least a dimensioned sketch of this tray, or whatever he was after a quote for, I'm afraid he'd be politely invited to fcuk off. I'm a busy man with a lot on at this time of year and I haven't got time to be working out what sizes of cakes he wants to sell. :E

BDiONU
19th Dec 2007, 20:00
Ok a seasonal one here, in the song The Twelve Days of Christmas how many gifts did the bloke get from his true love?
364

BD

Forkandles
19th Dec 2007, 20:09
364
...it goes like this
1st day = 1
2nd day = 3
3rd day = 6
4th day = 10
5th day = 15
6th day = 21
7th day = 28
8th day = 36
9th day = 45
10th day = 55
11th day = 66
12th day = 78
add that up and you get 364.

This is all to say that you get all the gifts stated every day. If you only get the gifts stated once you would get 78 total gifts.

Do I get a prize for copying this direct from google?

Brewster Buffalo
19th Dec 2007, 20:47
Having eaten a bit too much Christmas dinner you decide to go for a bit of a walk. You walk up the hill behind your house at 2mph (3.22 km/h). Then you walk back down it via the same route at 3mph (4.83 km/h).

What is your average speed?

BlooMoo
19th Dec 2007, 20:52
I was thinking as a kid would be expected to think...

You work for for the ALF?
:}

Right Way Up
19th Dec 2007, 21:19
BB,
2.4 mph ?
(x=distance, t1=time up, t2=timedown)
speed=dist/time

so journey up t1=x/2
journey down t2=x/3

Av speed = 2x/t1+t2
= 2x/(x/2 = x/3)
=12x/(3x+2x)
=12x/5x
=12/5 = 2.4

Arm out the window
19th Dec 2007, 21:29
I reckon 2.4, too.

PS - was working out the tray problem before, wife says "What are you doing?" - I say "errrr...maths?"
She gives me the look that says "You nerd!"

Ah well, there's a certain joy in maths, no matter what anyone says.

parabellum
19th Dec 2007, 21:53
"It is normal for any 'given' to be made explicit by the paper's author - eg "In answering you may assume that ....etc, etc,".

Bloo Moo - I think you may have misunderstood the state of education and the availability of competent examiners in the UK!;)

The usual examiner will have already been poached by an IT company and be working in the Bahamas whilst the regular Maths teacher will be otherwise engaged in "counselling" a group of pupils who got frightened by a big fluffy dog on the way to school and dropped their iPods, mobile 'phones etc. causing them to miss a pod cast from some useless teeeny band, or something equally serious.:confused:

IN the UK it would be quite common for the local baker to be setting the GCSE maths exam.;)

Brewster Buffalo
21st Dec 2007, 20:49
RWU & AOOTW - well done 2.4 is the right answer..

Next question -

You are making turkey stock for the Christmas day gravy. Overnight, part of the stock evaporates, making it go from being 99% water to 98% water. How much of the solution has evaporated?

The SSK
21st Dec 2007, 21:15
Why, half of it of course. But is it twice as tasty?

Brewster Buffalo
21st Dec 2007, 21:39
SSK - correct