Question

Find the dimensions of a rectangle with perimeter 84 m whose area is as large as possible. (if both values are the same number, enter it into both blanks.)

Answer 1

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.

Answer 2

All the sides of the rectangle are 21 m length.

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