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    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  (40x - 3x^2) m^2 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
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    What tells you the maximum or minimum points of a cubic equation?

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

    Spoiler:
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    differentiate it
    I don't think we've even done that yet :/
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    sketch the graph?
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    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.
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    (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
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    (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 :mad:

    For example, if you have the curve  y = x^2 + 3x - 10

    If you complete the square you get:

     y = (x + 1.5)^2 - 10 - 1.5^2

     y = (x + 1.5)^2 -12.25

    Now if you look at the (x + 1.5)^2 bit (the bit which changes value), because it is squared, then it will always be positive so the minimum value will be when (x + 1.5)^2 = 0 .

    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.
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    I know that completing the square can find the max/min point but have no idea how to apply it to 40x - 3x^2
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    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 =)
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    Okay through completing the square i've got...

     -3 (( x + \frac{20}{-3} )^2 - \frac{400}{9} )

     -3 (x + \frac{20}{-3} )^2 - \frac{400}{3}

    so the minimum point must be  \frac{400}{3} and that's the area?

    so for this equation there would only be one area? :confused:


    thankyou to everyone whos helped
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    ^ Already answered it for you.
 
 
 
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