Maths quiz

[Lucky]

I think he used the differentation method to get the length of a, assuming that the hypothenuse c is 3cm. With differentiation and b=5-a you´ll get a=2.5cm=b.
This would also satisfy the perimeter as 2*2.5+3=8.
And you would get an area of 2.5cm². That would be the maximum with this perimeter and one side being 3cm.
BUT a=b=2.5cm violates the Phytagoras equation: c²=a²+b². Therefore it would NOT be a right-angled triangle anymore!

That would be my explanation of this matter!

 

[ainwood]

Yep!

I think I know where AllHail was coming from, and I will make the wild assumption that I am right, and hence post the next question. Should be pretty easy, as its along the same lines as that other one.


You have a square of metal, 50cm x 50cm. You cut four equal squares out of the corners, and then fold up the sides to make a "tray". What is the length of each square that should be cut out to maximise the volume of the "tray".

 

[rjgo]

wow, ya'll are nerds.

get a life

shalom

rjgo

 

[ainwood]

Originally posted by rjgo
wow, ya'll are nerds.

get a life

shalom

rjgo

You're a college lad aren't you?

... If you have nothing constructive to add, please don't spam our thread.

 

[Hitro]

1. Volume = Ground * Height
V= a^2 * h (a^2 because it's a square)

2. 50 = a + 2h (one Ground side plus two heights)

=> a= 50 - 2h

V(h)= h * (50 - 2h)^2
= h*(2500 - 200h - 4(h^2))
= 2500h - 200(h^2) - 4(h^3)

V'(h)= 12(h^2) - 400h +2500
Now we need V'(h) = 0:
h^2 - (400/12)h +(2500/12) = 0

=> h = (200/12) + sqrt ( 10000/144) or (200/12) - sqrt(1000/144)
=> h = 25 or 8.3333 (= 100/12)
25 is nonsense cause in that case a would be = 0 and the whole thing wouldn't have a volume.

Therefore the correct answer is 8.333333333333 (= 100/12)
The max. volume is 9259,9259... cm^3

 

[ainwood]

Very good . Your turn

 

[Hitro]

Okay, I can' t think of anything intelligent now, so let's get on with some easy stuff.

Differentiate (once) this:

f(x) = (x^2 - 2) / (2x)

 

[Grisu]

f'(x)=(x^2+2)/2x^2

 

[Hitro]

EDIT: You are right Ovi.
Your turn.

 

[Grisu]

so let's do some geometry:

the points A,B,C together define the plain E (sorry I don't know if "plain" is the correct english term, it's Ebene in german, any help?). P* is the mirror-image of the point P mirrored on the plain E. What are the coordinates of the Point P*?
A = (3;-1;4) B = (8;2;1) C = (5;2;3) P = (25;2;29)

 

[Grisu]

hello ? anybody there???

c'mon this isn't that hard, is it?

 

[Lucky]

It should be plane!

Here is a nice English<->German online translator:
http://dict.leo.org .

Unfortunately I´m not in the mood right now to do this. I just finished 5 hours of work for my studies, so now is the time to do something else!

 

[Michiel de Ruyter]

Originally posted by KaeptnOvi
so let's do some geometry:

the points A,B,C together define the plain E (sorry I don't know if "plain" is the correct english term, it's Ebene in german, any help?). P* is the mirror-image of the point P mirrored on the plain E. What are the coordinates of the Point P*?
A = (3;-1;4) B = (8;2;1) C = (5;2;3) P = (25;2;29)

First, we need to define the plane (E) of A, B and C.
A to B gives (5,3,-3) and A to C (2,3,-1)
to get the equation for E we take the cross-produkt of these two vectors and get vector u =(6,-1,9)
Now we check it for A, B and C:
A : 6*3-1*-1+9*4=55
B: 6*8-1*2+9*1=55
C: 6*5-1*2+9*3=55
OK

Now to find the distance in u for P to E:
6(25+6x)-(2-x)+9(29+9x)=55
354+118x=0
-> x=-3
so the projection of P on E is (25-3*6 , 2+3 , 29-3*9)=P'=(7,5,2)
Now to check that: 6*7-1*5+9*2=55 -> correct
for the coordinates of P mirrored by E we just take 2*x = -6
and get (25-6*6 , 2+6 , 29-6*9)=P*=(-11,8,-25)

So I got (-11,8,-25)

 

[Grisu]

yep, absolutely correct.

your turn

 

[Michiel de Ruyter]

This is a series of numbers:

0,1,1,2,3,5

give the next five and the name of the creator

 

[anarchywrksbest]

0,1,1,2,3,5 - 8,13,21,34,55

Archimedes???

 

[AVN]

Fibonacci,

And the following numbers are 8,13,21,34,55

 

[Michiel de Ruyter]

Originally posted by AVN
Fibonacci,

And the following numbers are 8,13,21,34,55

You're right, it's Fibunacci, not Archimedes.
So you're next!

 

[AVN]

Again a series of numbers.

Which two numbers are following after :
11, 17, 23, 31, 41, 47, 59

and give a (short) explanation.

 

[Hitro]

Originally posted by AVN
Which two numbers are following after :
11, 17, 23, 31, 41, 47, 59

67 and 73

You took every second prime number, starting with 11, so after 59 comes 61 (left out) then 67, then 71 (left out) and then 73 .