Respuesta :
The dimensions of a rectangle with area 1728 square meter whose perimeter is as small(minimum) as possible are 41.57m and 41.56m.
Let the dimension of the rectangle be x and y m.
According to the given question.
The area of the rectangle is 1728 square meter.
⇒ x × y = 1728
⇒ x = 1728/y
Since, the perimeter of the rectangle is the sum of the length of its boundary.
Therefore,
Perimeter of the recatngle with dimensions x and y is given by
Perimeter of the rectangle, P = 2(x + y)
⇒ P = 2(1728/y + y)
⇒ P = 3456/y + 2y
Differentiate the above equation with respect to y
⇒ [tex]P^{'} = -\frac{3456}{y^{2} } + 2[/tex]
For the minimum(small) perimeter equate the above equation to 0.
⇒ [tex]-\frac{3456}{y^{2} } + 2 = 0[/tex]
⇒ [tex]2y^{2} =3456[/tex]
⇒ [tex]y^{2} = \frac{3456}{2}[/tex]
⇒ y^2 = 1728
⇒ y = √1728
⇒ y = 41.56 m
Therefore,
x = 1728/41.56
⇒ x = 41.57 m
Hence, the dimensions of a rectangle with area 1728 square meter whose perimeter is as small(minimum) as possible are 41.57m and 41.56m.
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