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

she will sew have sewn five dresses   and be 1/3 done with the sixth one in an hour.

step-by-step explanation:

3/4 of an hour=45 mins

lucy makes 1 dress per 11.25 minutes

45 mins+11.25 mins=56.25 mins

60 mins- 56.25 mins=3.75 mins

3.75 mins is 1/3 of of 11.25 mins

therefore she the other 1/3 would be part of the hour and she will have sewn 5 dresses and begun the sixth.

contact me if you have any questions you are welcome.