I'm really stuck with this problem. An isosceles triangle has sides of length x, x and p − 2x where p is the length of the perimeter of the triangle. Find the value of x which maximises the area of the triangle for fixed p and all the angles of the triangle for this value of x.
I get the area of the triangle as: but when I try to find the derivative I get a range of scary solutions that will not simplify. I only know about AS maths (AQA), the product rule and the chain rule, is there anything else I need? I know the answer is x=p/3 but I cannot get there...

melikecheese
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 25102015 19:08

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 25102015 20:14
(Original post by melikecheese)
I'm really stuck with this problem. An isosceles triangle has sides of length x, x and p − 2x where p is the length of the perimeter of the triangle. Find the value of x which maximises the area of the triangle for fixed p and all the angles of the triangle for this value of x.
I get the area of the triangle as: but when I try to find the derivative I get a range of scary solutions that will not simplify. I only know about AS maths (AQA), the product rule and the chain rule, is there anything else I need? I know the answer is x=p/3 but I cannot get there... 
ghostwalker
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 25102015 20:15
(Original post by melikecheese)
I'm really stuck with this problem. An isosceles triangle has sides of length x, x and p − 2x where p is the length of the perimeter of the triangle. Find the value of x which maximises the area of the triangle for fixed p and all the angles of the triangle for this value of x.
I get the area of the triangle as: but when I try to find the derivative I get a range of scary solutions that will not simplify. I only know about AS maths (AQA), the product rule and the chain rule, is there anything else I need? I know the answer is x=p/3 but I cannot get there...
Notice that p is a factor within your square root so you can take root p out.
After differentiating, you'll have a (...) ^ (1/2) some where. Pull it down into a denominator.../(...)^(1/2) and put everything over a common denominator.
You're then looking for the numerator to be zero, which should be a simple linear function.
If not, post working for someone to check. 
PrimeEpoch
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 25102015 22:19
If this is indeed a GCSE question, you do not need calculus. That's outside of the scope of the course, but by all means learn about it.
I tried doing this without calculus (yes there is a way, involving double roots and a lot of algebra), but I would have to use the cubic equation which is definitely out of the scope of GCSE. Turns out the easiest way is to just do calculus.
Since x can only be positive, f(x)^2=a^2, f(x)^2^(1/2)=a.
We know p/4<x<p/2 because of the roots of the equation. This will be our check, since we don't know the answer at this point (good practice for an exam).
I differentiated f(x)^2 such that d/dx [1/4 . (p2x)^2 (4pxp^2)] = 1/4[4p(p2x)^2  4(p2x)(4pxp^2)] = p(p2x)(p3x)
You probably know the rest, but I'll go through it anyway.
Since you want the turning points, you set dy/dx=0, so p(p2x)(p3x)=0, solving for x=p/2 and x=p/3. At the start, we said p/4<x<p/2, so x =/= p/2 for the maximum value (it's a root). This leaves x=p/3.
You can also tell by the shape of the curve. The coefficient of x^3 is +ve, so the curve comes from the bottom right quadrant, has a maximum, then a minimum, then goes back up to the top right quadrant. This makes the x coordinate of the minimum larger than the x coordinate of the maximum, and p/3 is smaller that p/2 (p is +ve), so the maximum must be p/3.
Hope this helps, I spent ages trying to do it within GCSE scope before I realised it wasn't even GCSE scope lol 
PrimeEpoch
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 25102015 22:27
Sorry, I somehow found this in the GCSE maths section, then again in A level maths. Please continue XD

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 25102015 22:32
(Original post by PrimeEpoch)
Sorry, I somehow found this in the GCSE maths section, then again in A level maths. Please continue XD
The 3 doors say:
GSCE Maths, A LEVEL Maths, Undergraduate Maths...
... but they all lead to the same room ... 
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 25102015 22:40
(Original post by TeeEm)
There is only one Maths room with 3 doors
The 3 doors say:
GSCE Maths, A LEVEL Maths, Undergraduate Maths...
... but they all lead to the same room ... 
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 25102015 22:42
(Original post by PrimeEpoch)
Weird. You can tell I'm new here lol 
melikecheese
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 26102015 20:12
(Original post by PrimeEpoch)
If this is indeed a GCSE question, you do not need calculus. That's outside of the scope of the course, but by all means learn about it.
I tried doing this without calculus (yes there is a way, involving double roots and a lot of algebra), but I would have to use the cubic equation which is definitely out of the scope of GCSE. Turns out the easiest way is to just do calculus.
Since x can only be positive, f(x)^2=a^2, f(x)^2^(1/2)=a.
We know p/4<x<p/2 because of the roots of the equation. This will be our check, since we don't know the answer at this point (good practice for an exam).
I differentiated f(x)^2 such that d/dx [1/4 . (p2x)^2 (4pxp^2)] = 1/4[4p(p2x)^2  4(p2x)(4pxp^2)] = p(p2x)(p3x)
You probably know the rest, but I'll go through it anyway.
Since you want the turning points, you set dy/dx=0, so p(p2x)(p3x)=0, solving for x=p/2 and x=p/3. At the start, we said p/4<x<p/2, so x =/= p/2 for the maximum value (it's a root). This leaves x=p/3.
You can also tell by the shape of the curve. The coefficient of x^3 is +ve, so the curve comes from the bottom right quadrant, has a maximum, then a minimum, then goes back up to the top right quadrant. This makes the x coordinate of the minimum larger than the x coordinate of the maximum, and p/3 is smaller that p/2 (p is +ve), so the maximum must be p/3.
Hope this helps, I spent ages trying to do it within GCSE scope before I realised it wasn't even GCSE scope lol 
PrimeEpoch
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 26102015 21:13
(Original post by melikecheese)
Thanks so much for this, it' really helpful! I just have 1 question, how did you get p/4<x<p/2 ?
You know as you increase the value of x, the area will start at 0 for the minimum possible value of x that can form a triangle, and will be 0 again at the maximum possible value of x that actually forms a triangle. Your solution must be in between these values of x, where a=0. That's the long winded way of saying it's in between the roots of your equation.
If you look at a^2, the roots are really obvious. Even if you look at a, you see that if cde=0, where c is a constant, d=p2x and e=(4px  p^2)^(1/2) or e=(4xp)^(1/2) depending if you took a factor of p out, you will notice that either d or e must equal 0 for the equation to hold. Obvious for d, but a little less obvious for e since you have a root over it. Once you equate it to 0, you can just take the root off and solve normally.
There's another way of thinking about it in case you're interested. You know that any one side of a triangle cannot be more than half of the total perimeter for the points to connect, and if any one side is half of the perimeter, then area=0. Your solution must be within the values x holds when it's in this state. if you take the side where length=x, you know that the length of the side is equal to half the perimeter therefore x=p/2. If you take the side where length=p2x, p2x=p/2, x=p/4. Then you know that x is between p/2 and p/4, so formally that's p/4<x<p/2.
Hope this helps 
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 26102015 21:23
(Original post by TeeEm)
There is only one Maths room with 3 doors
The 3 doors say:
GSCE Maths, A LEVEL Maths, Undergraduate Maths...
... but they all lead to the same room ... 
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 26102015 21:26
(Original post by djpailo)
Yes but why stop at having three doors. Why not more! 
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 26102015 21:29
(Original post by TeeEm)
why not 3 different rooms? 
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 26102015 21:34
(Original post by djpailo)
Just three?
one of the reasons GCSE student do not frequent TSR maths is that the single room is too intimidating.
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