Annoyingly hard HSC Question.. (1 Viewer)

[Dota55]

I can't seem to solve Q10 of this HSC Paper.

http://img239.imagevenue.com/img.php?image=68757_90_122_164lo.jpg

umm i think i got my ideas right...but i'm not confident that my answers are correct.

Also, i'm stuck on the second part.

Can someone please go through this question?

 

[sicmacao]

I can't seem to solve Q10 of this HSC Paper.

http://img239.imagevenue.com/img.php?image=68757_90_122_164lo.jpg

umm i think i got my ideas right...but i'm not confident that my answers are correct.

Also, i'm stuck on the second part.

Can someone please go through this question?

noice level from L1 is N1 = L1/x^2
N2 = L2/(m-x)^2

The noice level at P is N = N1+N2 = L1/x^2 + L2/(m-x)^2

dN/dx = -2L1/x^3 + 2L2/(m-x)^3
= 0
solving to get
L1/x^3 = L2/(m-x)^3

(note do not attempt to cross multiply and expand as the resulting cubic equation becomes difficult to solve.)

Instead rearrange to get the following
((m-x)/x)^3 = L2/L1

x = m/(cuberoot(L2/L1)+1)

Check minimum by 2nd derivative
d/dx(dN/dx) = 6L1/x^4 + 6L2/(m-x)^4 >0 for all x
(since L1, L2, x^4, (m-x)^4 are all positive)

since 2nd derivative is positive, means concave up, hence minimum noise at

x = m/(cuberoot(L2/L1)+1)

 

[JSBboag]

noice level from L1 is N1 = L1/x^2
N2 = L2/(m-x)^2

The noice level at P is N = N1+N2 = L1/x^2 + L2/(m-x)^2

dN/dx = -2L1/x^3 + 2L2/(m-x)^3
= 0
solving to get
L1/x^3 = L2/(m-x)^3

(note do not attempt to cross multiply and expand as the resulting cubic equation becomes difficult to solve.)

Instead rearrange to get the following
((m-x)/x)^3 = L2/L1

x = m/(cuberoot(L2/L1)+1)

Check minimum by 2nd derivative
d/dx(dN/dx) = 6L1/x^4 + 6L2/(m-x)^4 >0 for all x
(since L1, L2, x^4, (m-x)^4 are all positive)

since 2nd derivative is positive, means concave up, hence minimum noise at

x = m/(cuberoot(L2/L1)+1)

Thanks you so much, i tried to do it before but couldn't do it and then when i read your working it made it really easy to understand.

I just have one question, when you get the 1st derivative - i though L1 and L2 are co-effiecients and therfore should of used the quotient or product rule instead of the chain rule????

Thanks

 

[sicmacao]

L1 and L2 are constants, quotient rule and product rule does not apply
eg d/dx(4x^3) = 12x^2, d/dx(Lx^3) = 3Lx^2

 

[kokodamonkey]

if you want something hard.. try the "snow plough" one i think its question 10 from 2000. Thats the infamous hard one. apparantly only one kid in the state got it right.

 

[Dota55]

forgot to thank sicmacao for answering my question

sorry!

and thankyou !!

 

[IMABOYDAMON!]

if you want something hard.. try the "snow plough" one i think its question 10 from 2000. Thats the infamous hard one. apparantly only one kid in the state got it right.

I just showed the question to my brother, and he finished it in half an hour. LMAO!

 

[JSBboag]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics23U.pdf

help with the snow plough question

 

[lyounamu]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics23U.pdf

help with the snow plough question

Somebody already did it.

Go back to maths page and go to where it says "give you $20..."

 

[JSBboag]

Are you talking about the Maths report because they don't adequetly, systematically, and holistically answer the whole question.