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# C1 question watch

1. Basically I've got this shape

Question: A farmer has 40m of fencing and he decides to use the right angle corner of a building. Show that the area he can enclose is given by and deduce the maximum value of this area.

I get how to find the area but how do I get the maximum area from this when I don't know what x is?

thanks for any help
2. What tells you the maximum or minimum points of a cubic equation?

Spoiler:
Show
differentiate it
3. (Original post by Rubgish)
What tells you the maximum or minimum points of a cubic equation?

Spoiler:
Show
differentiate it
I don't think we've even done that yet :/
4. sketch the graph?
5. Just finished C1 and haven't done this.

Say you differentiate it and get 40 - 6x. Then what?

EDIT: Just realised that by making it equal to 0 you get x = 6 and 2/3.

Which is the value x must be to get the highest value.

Not entirely sure how that works but it does

Is it just like the max/min points on a curve? But in this case its the area of a field which is slightly confusing. Anyway I don't think max/min points of a curve are in C1.. they are in C3 I think along with the d^2y / dx^2 to see whether it is a min/max point.

Anyway, @ OP, to work out the maximum/minimum points on a curve it is where the gradient of dy/dx is equal to 0.
6. (Original post by Thrug)
Just finished C1 and haven't done this.

Say you differentiate it and get 40 - 6x. Then what?

EDIT: Just realised that by making it equal to 0 you get x = 6 and 2/3.

Which is the value x must be to get the highest value.

Not entirely sure how that works but it does

Is it just like the max/min points on a curve? But in this case its the area of a field which is slightly confusing. Anyway I don't think max/min points of a curve are in C1.. they are in C3 I think along with the d^2y / dx^2 to see whether it is a min/max point.

Anyway, @ OP, to work out the maximum/minimum points on a curve it is where the gradient of dy/dx is equal to 0.
Hmmm I have no idea how to differentiate but thanks I can get the answer from the 40-6x

I'm not sure why the questions under completing the square though o_O
7. (Original post by emma363)
I'm not sure why the questions under completing the square though o_O
Completing the square is another way for finding a maximum or minimum. My teachers never told me this until after C1

For example, if you have the curve

If you complete the square you get:

Now if you look at the bit (the bit which changes value), because it is squared, then it will always be positive so the minimum value will be when .

So in this case the minimum value occurs at x = -1.5, and the actual minimum value is 12.25 (the number left..) This works for all quadratics

Hope that makes sense.
8. I know that completing the square can find the max/min point but have no idea how to apply it to 40x - 3x^2
9. Assuming you've already been able to get the are to equal 40x-3x^2 the next thing you do is differentiate it:

f(x) = 40x-3x^2
f'(x) = 40-6x

You should know a quadratic graph has only one turning point on it (ie its a n shape because the x^2 term is negative) The maximum point is the very top of the n where the line starts to go back down - a turning point. Turning points have gradient of zero and knowing this all you have to do is equate your differentiated equation to zero:

once you've done the algebra you'll get a value for x (the variable involved with the side lengths) when y (the area) is at a maximum.

hope this helped =)
10. Okay through completing the square i've got...

so the minimum point must be and that's the area?

so for this equation there would only be one area?

thankyou to everyone whos helped

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Updated: October 3, 2010
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