Find the dimensions of a rectangle with area 1,331m² whose perimeter is as small as possible.

Solution:

Given, area of the rectangle is 1331 m²

We have to find the dimension of the rectangle whose perimeter is as small as possible.

Area of the rectangle = length × breadth

Let, x and y be the sides of the rectangle.

Area = xy

xy = 1331

y = 1331 / x -------- (1)

Perimeter = 2(x + y)

P = 2x + 2y -------- (2)

Substitute (1) in (2)

P = 2x + 2(1331 / x)

P = 2x + 2662 / x

Taking derivative,

P’ = 2 - 2662 / x²

To find critical points, P’ = 0

2 - 2662 / x² = 0

2 = 2662 / x²

x² = 2662 / 2

x² = 1331

Taking square root,

√x² =  √1331

x = 36.48m

Taking second derivative,

P’’ = 2(2662 / x³)

P’’ = 5324 / x³

5324 / x³ > 0

Thus, the perimeter will be minimum at x = 36.48 m.

Determine the second dimension,

y = 1331 / 36.48

y = 36.48 m

Therefore, the dimensions of the rectangle are x = y = 36.48 m.

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Find the dimensions of a rectangle with area 1,331m² whose perimeter is as small as possible.

Summary:

The dimensions of a rectangle with area 1,331m² whose perimeter is as small as possible are x = y = 36.48 m.