This is the questions................?

Question

A farmer wishes to build a rectangular sheep pen with a 160m fence,using one side of an existing shedas the side.The side is also rectangular of measurement 20 m x 20 m.

a)Find the measurement and the area of the sheep pen if one sideof the existing shed is used as the width of the sheep pen.

b)By using the fence of the same length and using the width of the shed as part of thewidth of the sheep pen,find the length and width of the sheep pen with maximum area.

c)If two sheep pen of the same size are built adjusted to each other,find
1)the length of the new sheep pen if one side of the shed is used as the width of the sheep pen.
(2)the length and width of the sheep pen,which is adjusted so that enclosed area are maximum.

d)Find the maximum enclosed area if the same fence is used to construct three rectangular sheep pens.

Further Ivestigation

If n rectangular sheep pens of width 20 m are constructed(one side of the existing shed is used as part of the pen)from a fence of length x m,find in term of x and n :

(1)the length of the sheep pen,
(2)total area of the sheep pens.

Answer 1

a)Since a side of shed is used no fence is req. there.
160 m is utilized for three remaining sides.
Since its a rectangle other side for widht is also fixed i.e.20 m
Thus remaining fence =160-20 = 140 m.
For the lenghts of the rectangle which are equal, divide 140 by 2
=70 m each
Thus measurement of sheep pen is 70 m x 20 m .

b)A rectangle having largest or maximum area is a square.
This sheep pen is hence a square.
Let x m be the extra lenght more than that of 20 m side.
Thus total lenght of the side of the square pen =20+x m
But since 20 m is used up by the side, lenght of fence to be used =x+3*(20+x) m
=4*x + 60
Lenght of fence=160 m
thus, 4x+60=160
4x=100
x=25
Lenght of side =20+x = 20+25= 45 m
Dimensions of pen =45m x 45m

.

c)1)From the conditions if x m is the lenght of rectangle
then 4x+2*20=160
x=120/4=30 m
Dimension of the pen is 30m x 20m

2)Consider the 2 pens adjacent lengthwise.
Thus,
60+3x=160
x=33.33m
The dimensions are 33.33m x 20m


d) assuming they are attached to the shed all of them are squares for max area.
6x+60=160
6x=100
x=16.66m
dimension of each one of three pens = 20m x 16.66m
Max area =333.33 sq m

Further invetigation : I didnot get the constraints properly.

Answer 2

Use the total area of 160m as one base, but add 20m to add in the shed.

Use the fact the shed is 20m as the other.

Then use your own maths skills to work it out.

Answer 3

sure