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2003 paper 2, question 8 - HSN forum

# 2003 paper 2, question 8

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### #1jambo david

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Posted 15 January 2006 - 08:29 PM

Hi, i'm having trouble with part b, trying to find the minimum surface area, i managed to do this for another question but it's just not working here, can anyone give me a hand. I was going to use www.mathsrevision.com but it's not working any more

### #2aldo

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Posted 15 January 2006 - 08:52 PM

Your starting point is to express the numerical value for volume, with a mathematical equation (otherwise your equation for surface area will not have such a large number in it).

So:

Volume = area of triangle x length of trough.

108 litres = (1/2) (x) (x) (l)

108 litres = (1/2) (x ) (l)

2 x 108 litres = (x ) (l)

216 litres = (x l)

And you should be aware that 216 litres = 216,000 ml.

The length can then be expressed as:

l = 216,000 / x

Now you can turn your attention to the surface area:

SA = 2 triangle areas + 2 rectangle areas

SA = 2 (1/2) (base) (height) + 2 (l) (b)

SA = (base) (height) + 2 (l) (b)

SA = x + 2 (216,000 / x ) (x)

SA = x + (432,000 x / x )

SA = x + (432,000 / x)

And this is the expression you are asked to obtain!!

### #3aldo

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Posted 15 January 2006 - 08:55 PM

Give me a few minutes and I'll work it through!!

### #4aldo

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Posted 15 January 2006 - 09:04 PM

Part (b) working in attached file (excluding the nature table, which you should do to show that at x = 60, the surface area is indeed a minimum).

### #5jambo david

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Posted 15 January 2006 - 09:29 PM

Haha, i just missed out the x^3, i had it quared. Anyway, i had the concept right but i'll definately make sure i don't make a mistake like that tomorrow. Cheers, really appreciate it

### #6xkarenx

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Posted 30 January 2006 - 09:56 PM

You did not waste your time typing out part a because I was confused by it and it helped me so thank you!

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