Respuesta :
The dimensions would be 21 m by 21 m.
To maximize area and minimize perimeter we use dimensions that are as close to equivalent as possible; 84/4 = 21, so we can use a square with side length 21.
To maximize area and minimize perimeter we use dimensions that are as close to equivalent as possible; 84/4 = 21, so we can use a square with side length 21.
Answer:
All the sides of the rectangle are 21 m length.
Step-by-step explanation:
The problem is:
Maximize: Area = a*b (eq. 1)
subject to: 2*a + 2*b = 84 m (eq. 2)
where a and b are the length of the sides of the rectangle.
Isolating a for eq. 2:
a = 84/2 - 2/2*b = 42 - b (eq. 3)
Replacing in eq. 1:
Area = (42 - b)*b = 42b - b^2
In the maximum the derivative is equal to zero. Then:
d Area/d b = 42 - 2*b = 0
42 = 2*b
b = 21 m
Replacing in eq. 3:
a = 42 - b = 42 - 21 = 21 m
