Find the dimensions of a rectangle with an area 1,000m2 whose perimeter is as small as possible.

Solution:

Let us consider a rectangle of Length X and Breadth Y :

The area of the rectangle (A) = XY m² = 1000m² --- (1)

The perimeter of the rectangle (P) = 2(X + Y)m --- (2)

The objective is to minimize the perimeter of the rectangle given the area of the rectangle. Therefore from equation (1) we can write,

Y = 1000/X --- (3)

Substituting equation (3) in (2) we get

P = 2(X + 1000/X)

P = 2X + 2000/X

To find the minimum area we need to differentiate the above equation w.r.t. X variable

dP/dX = 2 - 2000/X²

Equating it to zero we have

2 - 2000/X² = 0

2 = 2000/X²

X² = 1000

X = √1000

Y = 1000/X = 1000/√1000 = √1000

To verify that the perimeter P is minimum for the above calculated dimensions of X and Y we wtite the second order differential of the function P,

d²P/dX² = 4000/X³

Since X > 0 we can state d²P/dX² is +ve

Which suggests that the perimeter with calculated dimensions of X and Y is minimum.

The dimensions of a rectangle with an area 1,000m² whose perimeter is as small as possible.

Stated in other words that the rectangle of area 1000 m² is a square of side √1000 m,

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Find the dimensions of a rectangle with an area 1,000m² whose perimeter is as small as possible.

Summary:

The dimensions of a rectangle with an area 1,000m² whose perimeter is as small as possible is 10√10 and 10√10.