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# differeniation/integration watch

1. Can someone help me with this question, I'm pretty sure it to do with differeniation or integration, I can only do part i , I have attempted part ii but I think its the wrong technique to use.
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2. i. 2Pi.r+2x=400 --> x=200-Pi.r
ii. Area is x.2r. Just substitute for x here and you'll get the required result.
iii. dA/dr=400-4Pi.r=0 --> r=100/Pi
3. If I can read correctly what you've done for part (i) it looks wrong: you need to divide it by two as you have found 2x!

Then for part (ii) you take the correct answer for part (i) and multiply it by height: 2r. This gives the correct answer.

For part (iii) I believe you differentiate, giving 400-4πr
Set this equal to zero to find minima and maxima:
400-4pi.r=0
100/pi=r

x=200-pi.=200-100=100m
A=400(100/pi-2π(100/pi)²=40000/pi-20000/pi=20000/pi m²
4. Naystar: tick don't show smilies.
5. Oh I was using the ones when you "go advanced". And it was pi, not square root. Maybe I'll stick to writing pi.
6. (Original post by naystar)

x=200-pi.=200-100=100m
A=400(100/pi-2π(100/pi)²=40000/pi-20000/pi=20000/pi m²
Fo part iii i get r=31.8
x=100 and then A must be equal to x.2r ie width x length
or if you use A =400r-2πr² A=6366
Am I correct becasue you last line of working baffled me abit
7. I missd out closing a bracket- after 100/pi- sorry about that. You shouldn't round for r when calculating A until the very end, as you then end up with inaccuracies, but otherwise A would be correct. I think mine comes to 6369 or something like that, which is close so yours is correct.

And to find A I used A=400r-pi r²

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