Girls vs boys maths challenge Watch

Smaug123
  • Study Helper
Badges: 13
Rep:
?
#221
Report 5 years ago
#221
(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
0
reply
PhysicsKid
Badges: 0
Rep:
?
#222
Report 5 years ago
#222
The HCF is s, so is that what I factorise by, or do I do polynomial cubic division?
0
reply
Smaug123
  • Study Helper
Badges: 13
Rep:
?
#223
Report 5 years ago
#223
(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 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.
0
reply
PhysicsKid
Badges: 0
Rep:
?
#224
Report 5 years ago
#224
(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 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?
0
reply
PhysicsKid
Badges: 0
Rep:
?
#225
Report 5 years ago
#225
d/db = -s(s+a)(s+c)
d/dc = -s(s+a)(s+b)
So what do I need to do next?
0
reply
Smaug123
  • Study Helper
Badges: 13
Rep:
?
#226
Report 5 years ago
#226
(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 \dfrac{\partial f}{\partial b}, not \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 \dfrac{\partial F}{\partial k} for k=a,b and set them to 0 to find a turning point.
0
reply
PhysicsKid
Badges: 0
Rep:
?
#227
Report 5 years ago
#227
(Original post by Smaug123)
So that's \dfrac{\partial f}{\partial b}, not \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 \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?
0
reply
Dmon1Unlimited
Badges: 14
Rep:
?
#228
Report 5 years ago
#228
inb4 people use this to post their math homework

edit: seems ive already exposed the neggers intention for this thread
2
reply
Smaug123
  • Study Helper
Badges: 13
Rep:
?
#229
Report 5 years ago
#229
(Original post by PhysicsKid)
What's the difference between df/db and dF/db?
I defined F(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,\lambda) = s(s-a)(s-b)(s-c) + \lambda (s-\dfrac{a+b+c}{2}).
0
reply
PhysicsKid
Badges: 0
Rep:
?
#230
Report 5 years ago
#230
(Original post by Smaug123)
I defined F(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,\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)?
0
reply
Smaug123
  • Study Helper
Badges: 13
Rep:
?
#231
Report 5 years ago
#231
(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 a humble uni maths student, who would be considerably better at doing this in person rather than over the internet, but such is life…
0
reply
PhysicsKid
Badges: 0
Rep:
?
#232
Report 5 years ago
#232
(Original post by Smaug123)
Almost, but not quite - you've forgotten the minus sign from s-(that thing).I'm not a teacher 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
0
reply
PhysicsKid
Badges: 0
Rep:
?
#233
Report 5 years ago
#233
So dF/db = -s(s+a)(s+c)lambda(-1/2)?
0
reply
UniMastermindBOSS
Badges: 17
Rep:
?
#234
Report 5 years ago
#234
(Original post by Noble.)
Given that \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
0
reply
UniMastermindBOSS
Badges: 17
Rep:
?
#235
Report 5 years ago
#235
(Original post by Dmon1Unlimited)
inb4 people use this to post their math homework

edit: seems ive already exposed the neggers intention for this thread
lol that's actually a good idea (if I still had homework)
0
reply
Solemn Rain
Badges: 15
Rep:
?
#236
Report 5 years ago
#236
(Original post by Mr M)
4 \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!
1
reply
Solemn Rain
Badges: 15
Rep:
?
#237
Report 5 years ago
#237
there are hardly any girls here:rofl:
0
reply
Joel R
Badges: 7
Rep:
?
#238
Report 5 years ago
#238
(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 \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:
Show
y=6-2root3


?
1
reply
Smaug123
  • Study Helper
Badges: 13
Rep:
?
#239
Report 5 years ago
#239
(Original post by PhysicsKid)
So dF/db = -s(s+a)(s+c)lambda(-1/2)?
F(a,b,c,s,\lambda) = s(s-a)(s-b)(s-c) + \lambda (s-\dfrac{a+b+c}{2}), so \dfrac{\partial F}{\partial b} = -s(s-a)(s-c) - \lambda \times \dfrac{1}{2}. Do you see why?
0
reply
Technetium
Badges: 18
Rep:
?
#240
Report 5 years ago
#240
The discussions/ arguments on here are making me laugh!

I have a question: Show algebraically the value of the square root of i.
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

University open days

  • University of the Arts London
    MA Performance Design and Practice Open Day Postgraduate
    Thu, 24 Jan '19
  • Brunel University London
    Undergraduate Experience Days Undergraduate
    Sat, 26 Jan '19
  • SOAS University of London
    Postgraduate Open Day Postgraduate
    Sat, 26 Jan '19

Are you chained to your phone?

Yes (68)
19.43%
Yes, but I'm trying to cut back (144)
41.14%
Nope, not that interesting (138)
39.43%

Watched Threads

View All