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1. The dimensions of a rectangle (in m) with perimeter 84 m whose area is as large as possible or maximum area are 21 m and 21 m.

   Let the dimensions of the rectangle be x meters and y meters.

   According to the given question.

   The perimeter of the rectangle is 84m.

   ⇒ 84 = 2(x + y)

   ⇒ 42 = x + y

   ⇒ y = 42 – x

   As we know that the area of the rectangle is calulated by taking its product of length and width.

   So,

   Area of rectangle, A = xy

   ⇒ A = x(42 – x)

   ⇒ A = 42x – x^2

   Now, for the large possible area or maximum area, differentiate the above equation with respect to x and equate it to 0.

   ⇒ dA/dx = d(42x – x^2)/dx = 0

   ⇒ dA/dx = 42 – 2x = 0

   ⇒ 42 – 2x = 0

   ⇒ 42 = 2x

   ⇒ x = 21m

   Therefore,

   y = 42 – 21

   ⇒ y = 21m

   Hence, the dimensions of a rectangle (in m) with perimeter 84 m whose area is as large as possible or maximum area are 21 m and 21 m.

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