Respuesta :
Given :
Area of rectangle.
To Find :
The dimensions of a rectangle (in m) with area 1,728 m2 whose perimeter is as small as possible.
Solution :
Let, the dimensions of rectangle is x and y.
Area, A = xy.
x = A/y. ....1)
Perimeter, P = 2( x + y )
Putting value of x in above equation, we get :
[tex]P = 2( y + \dfrac{A}{y})[/tex]
For minimum P,
[tex]\dfrac{dP}{dx}=2( 1 - \dfrac{A}{x^2})=0\\\\A = x^2[/tex]
SO, it is a square.
[tex]x=\sqrt{A}\\\\x=\sqrt{1728\ m^2}\\\\x=24\sqrt{3}\ m[/tex]
Therefore, the dimensions are [tex](24\sqrt{3},24\sqrt{3})[/tex].24\sqrt{3}
Hence, this is the required solution.
The perimeter of the rectangle is the sum of its dimensions
The dimensions that minimize the perimeter are 41.6, 41.6
The area is given as:
[tex]\mathbf{A = 1728}[/tex]
Let the dimension be x and y.
So, we have:
[tex]\mathbf{A = xy = 1728}[/tex]
Make x the subject
[tex]\mathbf{x = \frac{1728}{y}}[/tex]
The perimeter is calculated as:
[tex]\mathbf{P = 2(x + y)}[/tex]
Substitute [tex]\mathbf{x = \frac{1728}{y}}[/tex]
[tex]\mathbf{P = 2(\frac{1728}{y} + y)}[/tex]
Expand
[tex]\mathbf{P = \frac{3456}{y} + 2y}[/tex]
Differentiate
[tex]\mathbf{P' = -\frac{3456}{y^2} + 2}[/tex]
Set to 0
[tex]\mathbf{ -\frac{3456}{y^2} + 2 = 0}[/tex]
Rewrite as:
[tex]\mathbf{ -\frac{3456}{y^2} = -2}[/tex]
Divide both sides by -1
[tex]\mathbf{\frac{3456}{y^2} = 2}[/tex]
Multiply y^2
[tex]\mathbf{3456= 2y^2}[/tex]
Divide by 2
[tex]\mathbf{1728= y^2}[/tex]
Take square roots of both sides
[tex]\mathbf{y = \sqrt{1728}}[/tex]
[tex]\mathbf{y = 41.6}[/tex]
Substitute [tex]\mathbf{y = \sqrt{1728}}[/tex] in [tex]\mathbf{x = \frac{1728}{y}}[/tex]
[tex]\mathbf{x = \frac{1728}{\sqrt{1728}}}[/tex]
[tex]\mathbf{x = \sqrt{1728}}[/tex]
[tex]\mathbf{x = 41.6}[/tex]
Hence, the dimensions that minimize the perimeter are 41.6, 41.6
Read more about perimeters at:
https://brainly.com/question/6465134
