The area of a rectangle whose perimeter is a fixed 80 feet is given by A=40w – w2 , where w is the width of the rectangle. Determine

The area of a rectangle whose perimeter is a fixed 80 feet is given by A=40w - w2 , where w is the width of the rectangle. Determine the width of the rectangle that gives the maximum area. What type of special rectangle is necessary to produce this maximum area? Justify.

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  1. Width=20 feet

    Since Length=Width=20 feet, the rectangle is a Square.

    Step-by-step explanation:

    Area, [tex]A=40w - w^2[/tex]

    To determine the width of the rectangle that gives the maximum area, we take the derivative of A and solve for its critical point.

    [tex]A'=40 - 2w\\$When A'=0\\40-2w=0\\40=2w\\w=20 feet[/tex]

    The width of the rectangle that gives the maximum area =20 feet.

    Perimeter of a rectangle=2(l+w)

    Perimeter of the rectangle=80 feet

    2(l+w)=80

    2l+2(20)=80

    2l=80-40

    2l=40

    l=20 feet

    Since the length and width are equal, the special type of rectangle that produces this maximum area is a Square.

  2. 25 questions , because they start with 100.  

    350-100=250 .  

    . 250/10=25

    step-by-step explanation:

    first take off 100, then what x 10 = 250? that is your answer. (25)

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