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2002 Winter Diet Calc. - Question 7 - help! - HSN forum

# 2002 Winter Diet Calc. - Question 7 - help!

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### #1Soph.E.

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Posted 19 May 2005 - 09:25 AM

This paper is just horrible, have worked out how to do the last few questions but still stumped on question 7! Here it is:

I know you'd do dS/dw and then find the maximum turning point but I can't get it to work out! And then finding d once you've got w... can anyone help?!

### #2dfx

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Posted 19 May 2005 - 11:15 AM

Yeah, same here. I've managed to differentiate, set it equal to zero, and got W = 200. Which doesn't really seem to make sense if the diameter is 20 only. And plus, how dyou get the value of d?

### #3Soph.E.

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Posted 19 May 2005 - 11:43 AM

Dear goodness, I have no idea. I thought these past papers were meant to get easier as the years go on but they seem to be getting more and more impossible! Maybe its just an over load of maths

I ended up with w equalling 11.5 but it's wrong

Did you get dS/dw = 680 - 51.w2?

### #4herbeey

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Posted 19 May 2005 - 12:02 PM

I get 680 = 5.1w^2...... I think...

W=133.333333333333333333333333333
That is what I am getting....
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### #5ermd

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Posted 19 May 2005 - 02:01 PM

QUOTE(herbeey @ May 19 2005, 12:02 PM)
I get 680 = 5.1w^2...... I think...

W=133.333333333333333333333333333
That is what I am getting....

You are right up to there. But it is W = 133.3333333333. So W = 133.3333333 = 11.55 to 2 dp.

Then sub into equation to find d? Edit: Maybe not, no d in the equation.

Surely the diameter must be inportant to this? Like to give the diagonals of the rectangle?

### #6herbeey

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Posted 19 May 2005 - 02:03 PM

Indeed. I think that is what I came up with.
However, once it equals that, how do you find 'd'?
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### #7ermd

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Posted 19 May 2005 - 02:08 PM

Right, because the diameter of the log is 20, radius is always 10 right. So from one corner to the centre of the rectangle is 10. Using pythagoras, you can get d now. But I don't know how to get it in exact value form.

### #8dfx

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Posted 19 May 2005 - 02:20 PM

Whoops my bad, sorry, it isnt w = 200. Was a surprisingly large figure. I've got the answer, posting in a sec.

### #9dfx

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Posted 19 May 2005 - 02:24 PM

Yeah, that's clever using Pythagoras, so obvious! lol.

I went to extreme lengths, modelled the circle relative to the coordinates x,y, with a centre (0,0)

So produces an equation, x^2 + y^2 = 100

Then since you know the total length of w = 11.55, take half of this (5.775), and this is where it intersects with the circle. So it intersects at the point x = 5.775. Plug it into the circle equation and get the value for the y intersection coordinate. And then simply double this value to get the whole length. Again, it isn't in exact value form, but calculators were allowed.

### #10gary

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Posted 19 May 2005 - 02:27 PM

Ok I got w=20 is this completely wrong.

### #11ermd

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Posted 19 May 2005 - 03:11 PM

I managed to get it in exact value form too . When gettin w, I just made sure to keep everything as exact values and it all worked out eventually.

### #12Soph.E.

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Posted 19 May 2005 - 03:17 PM

Ahhhh yes that makes tonnes more sense than what I was coming up with! In the answers it doesn't give the value for w (although I'm assuming it's 11.55) and I'm still having trouble getting the answer for d - it says 20 (2/3) and I'm nowhere near! Anyone feel like writing a quick working through once you've found w?! Thanks

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