Alan is building a garden shaped like a rectangle with a semicircle attached to one short side. if he 21) has 20 feet of fencing to go around it, what dimensions will give him the maximum area in the garden?

Let the short side be W

The half circumference of the semi-circle = .5 * pi *W

2Lengths + width + semicircle = 20 ft

2L + W + (.5*pi*W) = 20

2L + W + 1.57W = 20

Combine like terms

2L + 2.57W = 20

Simplify divide by 2

L + 1.285W = 10

L = (10 - 1.285W); for substitution:

Get the area equation:

Rectangle area + semicircle area

A = L * W + (.5*pi*(.5W) ^2)

A = LW + (1.57*.25W^2)

A = LW + .3925W^2

Replace L with (10-1.285W)

A = W(10-1.285W) + .3925W^2

A = -1.285W^2 + .3925W^2 + 10W

A = -.8925W^2 + 10W

Find W by finding the axis of symmetry of this equation a=-.8925, b=10

W = -10 / 2 * (-.8925)

W = -10 / -1.785

W = 5.60 is the width for max area

then we solve for the length

L = 10 - 1.285(5.60)

L = 10 – 7.20 = 2.8 ft is the length

Then, Check the perimeter

2(2.8) + 5.60 + 1.57(5.60) = 20ft

5.6 + 5.6 + 8.736 = 20ft

Rectangle: 2.8 by 5.60 has semicircle circumference 8.736 has a maximum area: (2.8*5.60) + (.5 * pi * 2.8 ^ 2)

=15.68 + 12.31

=28 sq/ft

a: 1/6, j: 1/4

step-by-step explanation: you are welcome

answer

false

explanation

these practices describe the manner the land use outcome is attained. land management methods implemented by corporations are a mandatory because they are a driver for better land use results economically, socially and environmentally. corporations are required to understand the connection between land use , and land management practices to support sustainable land use policies