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Thread: Quadratic Maximisation Questions Help

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    Quadratic Maximisation Questions Help

    Need help with question 9 and 10 please. Really struggling to get these questions right. Answers are attached below.

    9. A piece of wire 6 metres long is cut into two parts. One part is used to form a square and the other part is used to form a rectangle whose length is three times its breadth. Find the lengths of the two parts if the sum of the two areas is a maximum.

    10. A large open area is to have a section surrounded by a rectangular fence. This rectangle is then divided into six smaller rectangles, using one dividing fence parallel to its length and two fences parallel to its width. If the total length of fencing available is 1200 m, find the maximum possible area.
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    Last edited by 202025; 18 Feb 2019 at 6:29 PM.

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    Re: Quadratic Maximisation Questions Help

    Question 9:

    You need to convert the word problem into a number problem.

    Let the two pieces be x and y.

    First we know that the string is 6m long therefore
    X+Y=6
    So Y=6-X …Equation 1

    X is shaped into a square so each side is X/4 long and the area of the square is (X/4)^2

    Y is shaped into a rectangle with one side b and the other side 3b.
    Y is the perimeter so
    Y=b+b+3b+3b=8b
    So b=y/8
    The area of the rectangle is b*3b=3b^2
    So in terms of Y the area is 3*(y/8)^2
    And substituting in from Equation 1 the area of the rectangle becomes 3*((6-X)/8)^2

    We now have the two areas in terms of X

    A= (X/4)^2 +3*((6-X)/8)^2
    =(x^2)/16 + (3/64)(36-12X+x^2)

    The sum of the two areas now has to be maximised.

    A maximum occurs when
    dA/dX = 0
    =2X/16 + (3/64)(0-12+x^2)
    =X/8 +6X/64 -36/64

    (8X+6X)/64 =36/64
    14X=36
    X=36/14 =2 4/7

    Y=6-X = 3 3/7

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    Re: Quadratic Maximisation Questions Help

    Question 10:

    Let the length be L and the width be W.
    The fence goes around the outside, and there is one additional fence parallel with the length and 2 additional fences parallel with the width, and the total fence length is 1200m.

    Therefore
    1200=2L+L+2W+2W

    Rearranging:
    L=(1200-4W)/3

    The area of the rectangle
    A=LW
    =(1200-4W)/3 * W
    =400W – (4/3)W^2

    A maximum occurs when
    dA/dW = 0
    = 400 – 8W/3

    8W/3 = 400
    W=150

    Substituting:
    L=(1200-4*150)/3
    =200

    Area A=LW
    =150*200
    =30,000 m^2

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