Question Find the dimensions of a rectangle (in m) with perimeter 84 m whose area is as large as possible. (Enter the dimensions as a comma-separated list.)

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. Find out more information about possible large area or maximum area here: https://brainly.com/question/27233170 #SPJ4 Reply

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