Answer:
Area of rectangle = [tex]196\,m^2[/tex]
Length of rectangle = 14 m
Width of rectangle = 14 m
Step-by-step explanation:
Given:
Perimeter of rectangle is 56 m
To find: the maximized area of a rectangle and the length and width
Solution:
A function [tex]y=f(x)[/tex] has a point of maxima at [tex]x=x_0[/tex] if [tex]f''(x_0)<0[/tex]
Let x, y denotes length and width of the rectangle.
Perimeter of rectangle = 2( length + width )
[tex]=2(x+y)[/tex]
Also, perimeter of rectangle is equal to 56 m.
So,
[tex]56=2(x+y)\\x+y=28\\y=28-x[/tex]
Let A denotes area of rectangle.
A = length × width
[tex]A=xy\\=x(28-x)\\=28x-x^2[/tex]
Differentiate with respect to x
[tex]\frac{dA}{dx}=28-2x[/tex]
Put [tex]\frac{dA}{dx}=0[/tex]
[tex]28-2x=0\\2x=28\\x=14[/tex]
Also,
[tex]\frac{d^2A}{dx^2}=-2<0[/tex]
At x = 14, [tex]\frac{d^2A}{dx^2}=-2<0[/tex]
So, x = 14 is a point of maxima
So,
[tex]y=28-x=28-14=14[/tex]
Area of rectangle:
[tex]A=xy=14(14)=196\,m^2[/tex]
Length of rectangle = 14 m
Width of rectangle = 14 m
