2005 HSC Q8a (1 Viewer)

[shanks27]

Q- http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2005exams/pdf_doc/maths_ext2_05.pdf, Q8a

i understand u obv. have to f'(x) but in the worked solns in the maths. assoc book they take a step i just can't get. In their first line they go from
f(x) = (a+b+x)/3(abx)^1/3 to this
f'(x) = (1/3(ab)^1/3) . ((x^1/3 -1/3(a+b+x))/x^2/3) i can't seem to follow.

Anyone able to explain/ have an easier way i initially did it using the quotient rule but didn't get the right answer

 

[DrFunk]

I can't tell by looking at their step (sorry the computer layout makes it rough for me)
When I did it I used quotient rule, and from there did the usual max/min approach.

 

[ronnknee]

They simply factored out the constants and applied the quotient rule to make the process simpler

 

[Affinity]

Best way to do the question is to take logs first..


minimizing f is same as minimizing ln(f) = ln(a+b+x) - ln(x)/3 + ln(constant)

differentiating gives 1/[a+b+x] - 1/[3x]

not hard to get x = (a+b)/ 2 from there.

the 2nd derivative is
-1/[a+b+x]^2 + 1/(3x^2)

and if you substitute x = (a+b)/ 2 in there you get 8/9.. so it's a mininum

 

[Js^-1]

Best way to do the question is to take logs first..


minimizing f is same as minimizing ln(f) = ln(a+b+x) - ln(x)/3 + ln(constant)

differentiating gives 1/[a+b+x] - 1/[3x]

not hard to get x = (a+b)/ 2 from there.

the 2nd derivative is
-1/[a+b+x]^2 + 1/(3x^2)

and if you substitute x = (a+b)/ 2 in there you get 8/9.. so it's a mininum

When you get the ln f(x) = ...
and you differentiate, does the ln f(x) become f ' (x) / f (x)?
Because I got the answer, but its a bit shady...

 

[Affinity]

When you get the ln f(x) = ...
and you differentiate, does the ln f(x) become f ' (x) / f (x)?
Because I got the answer, but its a bit shady...

yes but that is not the point.. the argument goes:

Since ln is a monotonic, strictly increasing function, the x minimizing ln(f) is the same as the one which minimizes f.