Posted 01 December 2005 - 06:05 PM

**Q-** A rectangular field is 'x' metres long. It is enclosed by 200 metres of fencing.

**(a)** Find an expression in terms of 'x' for the breadth of the field

**(b)** Hence find the greatest area which can be enclosed with the 200 metres of fencing.

*Can someone tell me what I need to do, to solve this Q please

Posted 01 December 2005 - 07:04 PM

I'm not going to tell you the answer, but for part (a) you are told that the perimeter is 200 m and that the length is

*x* m.

Let

*b* be the breadth of the field, and then find a formula for the perimiter.

If you can't see how to go on from there, just post again.

Posted 02 December 2005 - 02:16 PM

I don't know how to do it, so can you help me.

Posted 02 December 2005 - 04:19 PM

Part (a) is probably S1/S2 level. Especially with that hint that George gave.

Posted 02 December 2005 - 05:05 PM

part a) the whole perimeter is 200m, so the total is 200. two sides we know is "x". Our job is to find the other side. So if we have 2 sides at length x what will the other two side be to give a total of 200

part b

Finding largest area is differentiation

Posted 02 December 2005 - 09:10 PM

It's actually differentiation.

Posted 02 December 2005 - 09:16 PM

OK, so we have a rectangle, whose perimeter is 200 m, and the length of one side is

*x* m.

The perimeter is given by

and you know that the length is

*x* m and you want to find the breadth, call it

*b* m.

So the perimeter is

m and you also know it's 200 m. So:

So the breadth in terms of

*x* is

metres. And that's part (a) done.

Posted 02 December 2005 - 10:17 PM

QUOTE(Steve @ Dec 2 2005, 09:10 PM)

It's actually differentiation.

och i thought it was differentiation but i somehow convinced myself it was integration

either way i have editted it on my post

Posted 03 December 2005 - 12:50 PM

**Q-** A curve is such that its derivative is defined as dy/dx= x - (1/x-squared)

If the curve passes through the point (2, -5/2), find the eqn of the curve.

*Do I sub in the x-coord into the dy/dx eqn. Then go onto solve, or is this wrong?

Posted 03 December 2005 - 12:58 PM

You use integration on the derivitive, then subsitute the x and y values into the integral to find C, then rewrite the equation with value for C which you just found

Posted 03 December 2005 - 06:22 PM

I intergrated the Eqn and I got:

1/2(x-squared) + x(power -1) + C

*Is this right dude? Btw how can I sub Y-values into it? After I sub X-values into it, what do I do?

Posted 03 December 2005 - 06:30 PM

because the equation you get when you integrate is equal to y, since when you integrate you need to remember to do both sides so integrating dy/dx gives you y

Posted 03 December 2005 - 06:36 PM

Posted 03 December 2005 - 06:41 PM

QUOTE(Steve @ Dec 2 2005, 10:16 PM)

OK, so we have a rectangle, whose perimeter is 200 m, and the length of one side is

*x* m.

The perimeter is given by

and you know that the length is

*x* m and you want to find the breadth, call it

*b* m.

So the perimeter is

m and you also know it's 200 m. So:

So the breadth in terms of

*x* is

metres. And that's part (a) done.

I then differentiated:

100x - (x-squared)

And it gave me x=100, is this correct?

Posted 03 December 2005 - 07:44 PM

Ignore me, i should learn to read

**Edited by John, 03 December 2005 - 09:01 PM.**

Posted 03 December 2005 - 08:28 PM

x - (1/x squared) dx

x- (x minus squared) dx

(x squared/2) - (x-1/-1) + C

1/2(x squared) + x(power -1)