Post Your Problem Here (2 Viewers)

[victorling]

Originally posted by Mill
Actually, I found a slightly quicker way.

Let f(x) = x * x^((x - Ln[x])/Ln[x])
= x^(x/Ln[x] - 1 + 1)
= x^(x/Ln[x])
= e^x

And again,
f'(x) = e^x


They're pretty much equivalent though.

well done!!!

 

[overwhelming]

Okay fine. I will give you a fairer question that I am working on.

Prove by Induction that the maximum number of regions created by n co-planar lines is 1/2.n(n + 1) + 1.

 

[KeypadSDM]

Originally posted by overwhelming
Prove by Induction that the maximum number of regions created by n co-planar lines is 1/2.n(n + 1) + 1.

What are you allowed to assume?

 

[overwhelming]

nothing.

 

[KeypadSDM]

Oh well that suckz0r.

Well if I could prove that the pattern of S_(n+1) = S_n + n + 1 Then I might be able to do it, but proving that pattern could prove a little tricky.

Seeing as you can't assume the pattern is correct, and I have NO idea how to prove it true.

 

[Grey Council]

uh, i think this really IS a two unit question.

I think overwhelming isn't saying it right. I think.

and well done Mills. Prior might be worth $5000 bucks a year after all. maybe. hehe

 

[overwhelming]

It is directly taken from a Mathematics text book.

 

[Grey Council]

was there a diagram?

and being a two unit question, i think it means over lapping sqaures. :S i've never done this type of question before. Wtf does co-planar mean?

 

[KeypadSDM]

co-planar means they lie in the same 2-d plane.

I.e. note how the x-axis splits the number plane into 2 parts. When you add the y-axis in , you get 4 parts. Now if you draw ANOTHER line in you can split it to a maximum of 7 parts.

 

[Seraph]

yes , i need help with approaching these type of questions.. cause i really get lost with them

A wire piece is 6m long it is cut into two parts , one is used to form a square and the other piece a rectangle whose length is three times its width. Find the length of the two parts if the sum of the area is a maximum.....

 

[victorling]

Originally posted by Seraph
yes , i need help with approaching these type of questions.. cause i really get lost with them

A wire piece is 6m long it is cut into two parts , one is used to form a square and the other piece a rectangle whose length is three times its width. Find the length of the two parts if the sum of the area is a maximum.....

This is a calculus question which involves differentiation of the sum of the areas of the square and the reactangle

Solution:

Divide the 6-m-wire into 2 parts, one with (6-x)m and another with x m

Let x m be the perimeter of the square, thus the four sides' length of the square would be (x/4) m respectively

Let (6-x) be the perimeter of the rectangle, since "length is 3times its width), we let the width be y and thus the length be 3y

Therefore the perimeter of the rectangle:
(6-x) = 2*(3y+y)

(6-x) = 8y

y = (6-x)/8 --------------------(1)


The area of the square: (x/4)*(x/4)

= (x*2)/16 metres square

The area of the rectangle: (3y)*(y)

= 3(y*2)

= 3[( (6-x)/8)*2] by(1)


Therefore the sum of the 2 areas:

= (x*2)/16 + 3[( (6-x)/8)*2]

Differentiation of this expression gives:

-36/64 + 14x/64---------(2)

When the sum of the area is a maximum,(2) = 0

Solving (2) gives you x = 24/7 m

Therefore one part is 24/7 m, another part is (6- (24/7)) m= 18/7 m

ANS: 24/7 m and 18/7 m

This question can also be solved by letting the side of the square be(6-x/4) and the width be (x/8), which would give u the same answers

 

[Seraph]

umm for the sum of the two areas i ended up with

(3x^2 - 36x + 108/192) + (x^2/16)

hmmm.. wot did i do wrong?


argh wot the hek why does my teacher give me crap out of New SEnior Maths!!!!

 

[KeypadSDM]

If you can solve the hard problems, you'll be able to kill the easy ones no problem.

That's probably why.

 

[Grey Council]

new senior maths is good. Whats wrong with it?

 

[victorling]

Originally posted by Seraph
umm for the sum of the two areas i ended up with

(3x^2 - 36x + 108/192) + (x^2/16)

hmmm.. wot did i do wrong?


argh wot the hek why does my teacher give me crap out of New SEnior Maths!!!!

"(3x^2 - 36x + 108/192)" is wrong defintely

 

[Seraph]

im not sure how its wrong though

b = (X-6)/8

So 3b = 3(x-6/8)
3b = (3x-18)/24

then 3b x b is the area of the 2nd rectangle so
(3x-18)/24 x (x-6)/8
(3x^2 - 18x - 18x + 108) / 192
= (3x^2 - 36x + 108) / 92

 

[victorling]

reply to Seraph


b= (6-x) NOT (x-6)

 

[victorling]

because the length of the whole wire is 6m
therefore if we let one part of the wire be x, then the remaining part must be 6-x

 

[Grey Council]

lol, bad luck! silly mistake, seraph.

 

[Seraph]

Um....

....

.....
Crap? !!!

thx