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.)

1. niczorrrr
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.