C2 application of differentiation Qs

Could somebody please help me out with these questions? They're from the C2 book, ex. 9C.

1) A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle. Given that the total length of the fence is 80 m, show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden. Given that the area is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.

3) A sector of a circle has area 100 cm^2. Show that the perimeter of this sector is given by the formula P=2r + SQRT100/Pi. Find the maximum value for the perimeter of such a sector. (we always use radians, right??)

4) A shape consists of a rectangular base with a semicircular top, as shown. Given that the perimeter of the shape is 40cm, show that its area, A cm^2, is given by the formula A=40r - 2r^2 – Pi r^2 / 2, where r cm is the radius of the semicircle. Find the maximum value for this area.

Reply 1

Christophicus

I did these recently. I'll get you started on each.

1) A rectangular garden is fenced on three sides, and the house forms the fourth side of the rectangle. Given that the total length of the fence is 80 m, show that the area, A, of the garden is given by the formula A=y(80-2y), where y is the distance from the house to the end of the garden. Given that the area is a maximum for this length of fence, find the dimensions of the enclosed garden, and the area which is enclosed.

Let the length of the fence between the house and to the end of the garden be y and the other side of the fence, x. Imagine a rectangular shape where you are only considering three sides.

Therefore;
2y + x = 80
x = 80 - 2y

Area of garden = xy = y(80 - 2y)

Area is a maximum, therefore dA/dy = 80 - 4y = 0 at maximum.
80 = 4y
y = 80/4 = 20

x = 80 - 2y = 80 - 2(20) = 40

Dimensions
40m by 20m

Area = 40*20 = 800m^2

3) A sector of a circle has area 100 cm^2. Show that the perimeter of this sector is given by the formula P=2r + SQRT100/Pi. Find the maximum value for the perimeter of such a sector. (we always use radians, right??)
Area of sector
0.5r^2*theta = 100
r^2*theta = 200
theta = 200/r^2

Perimeter = r + r + r*theta (r*theta = length of the arc of the sector)
Perimeter = 2r + r(200/r^2) = 2r + 200/r

You typed out the question wrong. I have the book in front of me.

the reason for r being greater than root(100/pi) is that the maximum size odf the sector can be a full circle.
If this is so,
the area = pi*r^2 = pi*(root(100/pi))^2 = 100pi/pi = 100cm^2
If the sector is smaller, the radius has to be greater than this value, as pi and the area (100cm^2) are constants.

And yes, you always use radians in sector area calculations.

Minimum value for the perimeter of such a sector
dP/dr = 2 - 200/r^2 = 0
200/r^2 = 2
r^2 = 200/2 = 100
r = 10

P = 2r + 200/r = 2(10) + 200/10 = 20 + 20 = 40cm

4) A shape consists of a rectangular base with a semicircular top, as shown. Given that the perimeter of the shape is 40cm, show that its area, A cm^2, is given by the formula A=40r - 2r^2 – Pi r^2 / 2, where r cm is the radius of the semicircle. Find the maximum value for this area.

Let the horizontal distance shown in the diagram between the edge of the shape and the edge of the semicircle be x. The vertical height of shape is 2r.

Perimeter = 2r + 2x + r*pi = 40
x = (40 - 2r - pi*r)/2

Area of shape
= 2r*x + 0.5r^2*theta
= 2r[20 - r - 0.5pi*r] + 0.5r^2(pi) (theta = pi radians)
= 40r - 2r^2 - pi*r^2 + 0.5pi*r^2
= 40r - 2r^2 - (pi*r^2)/2

Maximum value of area
dA/dr = 40 - 4r - pi*r = 0
r(4 + pi) = 40
r = 40/(4 + pi)

Area
= 40r - 2r^2 - 0.5pi*r^2
= 40(40/(4+pi)) - 2(40/(4+pi))(40/(4+pi)) - 0.5pi(40/(4+pi))(40/4+ pi))
= 40/(4+pi)[40 - 80/(4+pi) - 20pi/(4+pi)
= 40/(4+pi)[20]
= 800/(4 + pi)cm^2