Question

Find the dimensions of a rectangle with area 512 m2 whose perimeter is as small as possible. (If both values are the same number, enter it into both blanks.)

Answer 1

Answer:

√512 by √512

Step-by-step explanation:

Length the length and breadth of the rectangle be x and y.

Area of the rectangle A = Length * breadth

Perimeter P = 2(Length + Breadth)

A = xy and P = 2(x+y)

If the area of the rectangle is 512m², then 512 = xy

x = 512/y

Substituting x = 512/y into the formula for calculating the perimeter;

P = 2(512/y + y)

P = 1024/y + 2y

To get the value of y, we will set dP/dy to zero and solve.

dP/dy = -1024y⁻² + 2

-1024y⁻² + 2 = 0

-1024y⁻² = -2

512y⁻² = 1

y⁻² = 1/512

1/y² = 1/512

y²  = 512

y = √512 m

On testing for minimum, we must know that the perimeter is at the minimum when y = √512

From xy = 512

x(√512) = 512

x = 512/√512

On rationalizing, x = 512/√512 * √512 /√512

x = 512√512 /512

x = √512 m

Hence, the dimensions of a rectangle is √512 m  by √512 m