Find the length and width of a rectangle that has the given area and a minimum perimeter. Area: 32 square feet

Answer 1

for a given area, minimum perimeter would be with a square.

A=s^2 = 32 where s= a side of the square

s=sqr32 = 4sqr2

length=width = 4sqr2 = about 4(1.414) = about 5.66 feet

perimeter =4s = 4(4sqr2) = 16sqr2 = about 16(1.414) = 22.6

A=LW

32=LW

L=32/W

or using calculus

Perimeter = P = 2L+2W = 2(32/W) + 2W = 64/W + 2W

take the derivative and set equal to zero, then solve for W

P'(W) = 2 -64/W^2 = 0

2W^2 =64

W^2 = 32

W =sqr32 = 4sqr2 = width of the rectangle that minimizes perimeter

L= 32/sqr32 = sqr32 = 4sqr2 = length of the rectangle that minimizes perimeter

L=W means a square with 4 sides each = 4sqr2

Answer 2

Let x = the length and y = the width.

Given: Area = xy = 32 --> y = 32/x

Perimeter P = 2x+2y = 2x+64/x

P'(x) = 2-64/x² = (2x²-64)/x²

To find a critical value, set the numerator to zero and solve for x

2x²=64

x=±√32 = ±4√2 take the positive value

y = 32/x = 32/(4√2) = 8/√2 = 4√2

This is a square with both the width and length = 4√2 feet.