Question

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?

Answer 1

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