The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

Girls vs boys maths challenge

Scroll to see replies

Original post by PhysicsKid
So how could I do that with my expanded form?


Your expanded form is quite right - you just need to factorise it again. Our answers are the same, but yours is a bit longer and a bit less useful when it comes to solving the simultaneous equations we end up with :smile:
The HCF is s, so is that what I factorise by, or do I do polynomial cubic division?
Original post by PhysicsKid
The HCF is s, so is that what I factorise by, or do I do polynomial cubic division?

Factorise out the HCF first - polynomial factorisation (for anything bigger than quadratic) is something of a dark art, but fortunately there's this nice factor of s to pull out :smile: Then you'll have a quadratic, and it's easy to factor a quadratic by finding its roots using the quadratic formula or completing the square or intuition or something.
Original post by Smaug123
Factorise out the HCF first - polynomial factorisation (for anything bigger than quadratic) is something of a dark art, but fortunately there's this nice factor of s to pull out :smile: Then you'll have a quadratic, and it's easy to factor a quadratic by finding its roots using the quadratic formula or completing the square or intuition or something.


I've taken out -s to give:
-s(s^2+bs+cs+bc)
The quad can be written:
s^2+(b+c)s+bc
Which means that the factorisation is:
(s+b)(s+c)
The factorised form is:
-s(s+b)(s+c)
What do you do next?
d/db = -s(s+a)(s+c)
d/dc = -s(s+a)(s+b)
So what do I need to do next?
Original post by PhysicsKid
d/db = -s(s+a)(s+c)
d/dc = -s(s+a)(s+b)
So what do I need to do next?


So that's ∂f∂b\dfrac{\partial f}{\partial b}, not ∂F∂b\dfrac{\partial F}{\partial b}. You really want the latter, because that's the expression that's got the constraint built in.
Then you work out ∂F∂k\dfrac{\partial F}{\partial k} for k=a,b and set them to 0 to find a turning point.
Original post by Smaug123
So that's ∂f∂b\dfrac{\partial f}{\partial b}, not ∂F∂b\dfrac{\partial F}{\partial b}. You really want the latter, because that's the expression that's got the constraint built in.
Then you work out ∂F∂k\dfrac{\partial F}{\partial k} for k=a,b and set them to 0 to find a turning point.


What's the difference between df/db and dF/db?
Original post by PhysicsKid
What's the difference between df/db and dF/db?

I defined F(a,b,c,s,λ)=f(a,b,c,s)+λ×somethingF(a,b,c,s,\lambda) = f(a,b,c,s) + \lambda \times \text{something}, where the something represents the constraint.
So f(a,b,c,s) = s(s-a)(s-b)(s-c), while F(a,b,c,s,λ)=s(s−a)(s−b)(s−c)+λ(s−a+b+c2)F(a,b,c,s,\lambda) = s(s-a)(s-b)(s-c) + \lambda (s-\dfrac{a+b+c}{2}).
Original post by Smaug123
I defined F(a,b,c,s,λ)=f(a,b,c,s)+λ×somethingF(a,b,c,s,\lambda) = f(a,b,c,s) + \lambda \times \text{something}, where the something represents the constraint.
So f(a,b,c,s) = s(s-a)(s-b)(s-c), while F(a,b,c,s,λ)=s(s−a)(s−b)(s−c)+λ(s−a+b+c2)F(a,b,c,s,\lambda) = s(s-a)(s-b)(s-c) + \lambda (s-\dfrac{a+b+c}{2}).


Thanks so much for you help? Are you a teacher, you should be? So when I differentiate lambda(s-(a+b+c)/2) for dF/db, do I get lambda(1/2)?
Original post by PhysicsKid
Thanks so much for you help? Are you a teacher, you should be? So when I differentiate lambda(s-(a+b+c)/2) for dF/db, do I get lambda(1/2)?


Almost, but not quite - you've forgotten the minus sign from s-(that thing).
I'm not a teacher :smile: a humble uni maths student, who would be considerably better at doing this in person rather than over the internet, but such is life…
Original post by Smaug123
Almost, but not quite - you've forgotten the minus sign from s-(that thing).I'm not a teacher :smile: a humble uni maths student, who would be considerably better at doing this in person rather than over the internet, but such is life…
So it's lambda (-1/2)? And yeah trying to type and learn latex and stuff is really annoying- I am SO much faster writing on paper or my whiteboard :smile:
(edited 10 years ago)
So dF/db = -s(s+a)(s+c)lambda(-1/2)?
Original post by Noble.
Given that xy2\dfrac{x}{y^2} is a minimum you know exactly where the centre of the two circles will be.


should I know this? I'm staying out of this thread :frown:
lol that's actually a good idea (if I still had homework)
Original post by Mr M
434 \sqrt 3

Can I answer any more or am I frozen out of the game?


:lol: You're clearly an expert on the subject! No, you should continue! GOGO!
there are hardly any girls here:rofl:
Reply 236
Avatar for Joel R
Joel R
7
Original post by Mr M
I think the winner of the previous round should set the next question. This is secondary school level but reasonably difficult. You won't find it anywhere on the internet as I made it up so there is no point Googling. I do have a worked solution.

Question 2

A scalene right-angled triangle with perimeter x cm and area x cm^2 contains a pair of congruent semicircles of radius y cm that do not overlap.

Given that xy2\displaystyle \frac{x}{y^2} is a minimum, find the exact value of y.


Apologies if this has already been answered, but I spent so long at such a late hour trying to get the answer I can't be bothered surfing through the whole thread.

Spoiler



?
Original post by PhysicsKid
So dF/db = -s(s+a)(s+c)lambda(-1/2)?


F(a,b,c,s,λ)=s(s−a)(s−b)(s−c)+λ(s−a+b+c2)F(a,b,c,s,\lambda) = s(s-a)(s-b)(s-c) + \lambda (s-\dfrac{a+b+c}{2}), so ∂F∂b=−s(s−a)(s−c)−λ×12\dfrac{\partial F}{\partial b} = -s(s-a)(s-c) - \lambda \times \dfrac{1}{2}. Do you see why?
The discussions/ arguments on here are making me laugh!

I have a question: Show algebraically the value of the square root of i.
Original post by Technetium
The discussions/ arguments on here are making me laugh!

I have a question: Show algebraically the value of the square root of i.


Oooh oooh pick me!

Spoiler

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you